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A three dimensional set of limit points related to the abc Conjecture Simpson, Reginald M. 2017

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A Three Dimensional Set of Limit Points Related to the abcConjecturebyREGINALD M. SIMPSONA THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THEREQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinTHE FACULTY OF GRADUATE AND POSTDOCTORAL STUDIES(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)December 2017c© Reginald M. Simpson, 2017AbstractWe study the limit points Q′ of a three-dimensional set Q which encodes the reciprocal qualityof abc triples as the components of a vector of the form (log Rad a, log Rad b, log Rad c)/ log c. Weestablish that if the abc Conjecture holds, Q′ is contained in a heptahedron. Unconditionally, weestablish the existence of a subset of Q′ with non-zero measure. We determine the implications ofprevious research on related problems involving limit points of abc triples in one-dimensional setson Q′ and discuss possible avenues for future study.iiLay SummaryThis thesis determines some of the mathematical properties of a three dimensional shape thatrepresents the factorization of triples of positive whole numbers (a, b, c) where a + b = c. Inparticular, this work builds on the work of researchers who studied a related problem involving aone-dimensional shape, and proves some new results about the three dimensional shape.iiiPrefaceThis thesis is comprised almost entirely of original, unpublished work that I wrote. Though I wrotethe text, chapters 4 through 6 contain original research done jointly with Greg Martin, for whichwe take equal credit. Both Julia Gordon and Greg Martin contributed to the editing of this work.Two published results from other researchers are quoted in this thesis. They are the BFGSTheorem from the beginning of Section 3 and Lemma 6.13. These two mathematical results arethe only unoriginal, previously published material in this thesis.The subject of this thesis was suggested by Greg Martin. Though it has not, as of December2017, been submitted for publication anywhere, Greg Martin and I have plans to prepare some ofthe material from Sections 4 to 6 for publication.ivTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 The Study of the Limit Points of Q with Polynomials and Binary Forms . . . . . . . . . . 53.1 Translation of the Results of [1] to Three Dimensions . . . . . . . . . . . . . . . . . . 63.2 Applying the BFGS Theorem to Other Binary Forms . . . . . . . . . . . . . . . . . . 123.3 Examining Q′ ∩ ([0, 1]3 \ H) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.4 Limitations of Current Polynomial Methods . . . . . . . . . . . . . . . . . . . . . . . 163.5 Related Limit Points and Proof Limitations . . . . . . . . . . . . . . . . . . . . . . . 174 Technical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.1 Sets of abc and almost abc triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 The G(q) function and related functions . . . . . . . . . . . . . . . . . . . . . . . . . 215 The Trivial Point Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Determination of a Volume in Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.2 Evaluation of U(N), and Bounds For Uδ and F . . . . . . . . . . . . . . . . . . . . . 376.3 Major Arc Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.4 Bounds of Functions on Component Arcs . . . . . . . . . . . . . . . . . . . . . . . . 566.5 Triple ∆ Error Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606.6 Single f?, Double ∆ Error Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646.7 Harmonic Arc Error Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.8 Double f?, Single ∆ Error Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77v6.9 Computation of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.10 Proof of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 927.1 Binary Forms, Polynomial Identities, and Limit Points in Three Dimensions . . . . . 927.2 Circle Method, Possible Optimizations and Generalizations . . . . . . . . . . . . . . 937.3 Relationship with Kane’s Paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98A Intersection of a Plane and Rectangular Prism . . . . . . . . . . . . . . . . . . . . . . . . 99viAcknowledgementsI would like to thank both my advisors, Julia Gordon and Greg Martin, for their support andguidance throughout my studies. I would also like to thank my mother, Carol Simpson, for hersupport of my education, and for her love.viiChapter 1IntroductionIn the study of the abc conjecture, the quality q of an element (a, b) ∈ N2 with gcd(a, b) = 1 isdefined to beq(a, b) =log(a+ b)log Rad ab(a+ b),where Radn denotes the radical (the largest squarefree divisor) of the integer n. In fact, oneformulation of the abc conjecture is expressed solely in terms of quality [1, 3].The abc Conjecture (Limit Point Formulation). The set S = {q(a, b) : a, b ∈ N, gcd(a, b) = 1}has a maximum limit point of 1. In other words, supS′ = 1.This set of limit points has been studied by other researchers, and some interesting conclusionshave been drawn. A paper by Browkin, Filaseta, Greaves and Schinzel [1] proves:1. Unconditionally, [1/3, 15/16] ⊂ S′ [1, Theorem 1].2. If the abc conjecture holds, S′ = [1/3, 1] [1, Theorem 4].Similarly, in a paper by Filaseta and Konyagin [3] it was shown thatS′ ∩ [1, 3/2) 6= ∅.In this work, we shall investigate a three-dimensional analog of S, which shall be called Q ⊂ R3.To define this three-dimensional analogue, we shall start with the inverse quality iq(a, b) = q(a, b)−1.First, we define−→iq(a, b) =1log(a+ b)(log Rad a, log Rad b, log Rad(a+ b)).1Observe that this is a natural way of splitting iq(a, b). Hence for−→iq(a, b) = (x, y, z), x + y + z =iq(a, b). iq =−→iq(a, b) · (1, 1, 1). This gives us our set:Q = {−→iq(a, b) : a, b ∈ N, gcd(a, b) = 1}.We shall denote by H ⊂ R3 the heptahedron enclosed by the seven planesx = 0, x = 1, y = 0, y = 1, z = 0, z = 1, x+ y + z = 1.Proposition 1.1. If the abc conjecture holds, Q′ is a subset of H.Proof. 1. First, we prove that if (α, β, γ) ∈ Q′, then α+ β + γ ≥ 1.If (α, β, γ) ∈ Q′, it follows that there exists a sequence {(an, bn)}∞n=1 with gcd(an, bn) =1 for all n and limn→∞−→iq(an, bn) = (α, β, γ). As (x, y, z) 7→ x + y + z is continuous,limn→∞ iq(an, bn) = α + β + γ. Thus if α + β + γ 6= 0, then (α + β + γ)−1 ∈ S′ and ifα+ β + γ = 0, then ∞ ∈ S′.If the abc conjecture holds, supS′ = 1, so it follows∞ 6∈ S′ and for α+β+γ 6= 0, (α+β+γ)−1 ∈S′ implies α+ β + γ ≥ 1. Thus it follows that if (α, β, γ) ∈ Q′, and the abc conjecture holds,then α+ β + γ ≥ 1.2. Second, we observe that trivially from the definition Q ⊆ [0, 1]3.Thus Q ⊂ [0, 1]3, and thus Q′ ⊂ [0, 1]3.Hence, by the definition of H, we have that H contains Q′.In other words, if the abc conjecture holds, H contains all possible limit points of Q. In thiswork, we seek to explore the shape and structure of Q′.2Chapter 2Summary of ResultsOur first approach, which is detailed in §3, will attempt to apply the same techniques as foundin [1] to prove the existence of points in S′, so as to determine the existence of points in Q′.For the following results, the notation (x1, y1, z1) → (x2, y2, z2) denotes a closed line segmentwith endpoints (x1, y1, z1), (x2, y2, z2). We prove that:Theorem 2.1. Unconditionally, the following hold:• If (x, y, z) ∈ Q′, then (y, x, z) ∈ Q′.• For n = 1, 2, 4, 6, the line segments ( 1n , 1, 1n )→ (0, 1, 1n ) are subsets of Q′.• For n = 1, 2, 4, 6, the line segments (1, 1n , 1n )→ (1, 0, 1n ) are subsets of Q′.• The line segments ( 34 , 12 , 13 )→ ( 14 , 12 , 13 ) and ( 12 , 34 , 13 )→ ( 12 , 14 , 13 ) are subsets of Q′.• The line segments ( 25 , 720 , 720 )→ ( 25 , 1960 , 720 ) and ( 720 , 25 , 720 )→ ( 1960 , 25 , 720 ) are subsets of Q′.• For n = 1, 2, 3, the line segments ( 12 + 12n , 12 , 12n )→ ( 12 , 12 , 12n ) are contained in Q′.• For n = 1, 2, 3, the line segments ( 12 , 12 + 12n , 12n )→ ( 12 , 12 , 12n ) are contained in Q′.Furthermore, if the abc conjecture holds, then the following holds:• For n ∈ N, t ∈ [0, 1] the point ( tn , 1n , 1) lies in Q′ or the point ( tn , 1, 1n ) lies in Q′.• The points (0, 1, 0), (1, 0, 0), and (0, 0, 1) are in Q′.Proof. First, we see the first statement is clear:3Observe that if (x, y, z) ∈ Q′ there exists a sequence of {(an, bn)}∞n=1 ⊆ N2, such thatlimn→∞−→iq(an, bn) = (x, y, z).Thus it follows that limn→∞−→iq(bn, an) = (y, x, z), and thus (y, x, z) ∈ Q′. This completes theproof of our first statement.The remaining statements follow from various lemmas and the symmetry implied by the firststatement. The unconditional statements are proven in Lemmata 3.1, 3.2, 3.2, and 3.5. Whenassuming the abc conjecture, we have the remaining statements by Lemma 3.4 and that Q′ is aclosed set so one may take 1/n→ 0.As is expected, all of the above statements, when projected back onto one dimension, give usthe statements found in [1]. What is of interest, however, is that while the results of [1] almostcompletely characterize the set S′, the same techniques when applied to the study of the three-dimensional object Q′ give only disjoint line segments.It is also known as a consequence of [3] that there exists a point on the plane x+ y + z = 1 inQ′ or below it. However, the technique cannot be easily extended to three dimensions. See §3.3 fordetails.After some examination of what it would take to modify the results of [1] to gain access to avolume, it was decided to approach the problem using other methods. The method of choice, in thiscase, being the circle method, with specific techniques taken from a paper [2] by Bru¨dern, Granville,Perelli, Vaughan, and Wooley. The application of this method to this problem is non-trivial, but avolume was obtained.Theorem 2.2. The volume enclosed by the following planes lies in Q′:0.7α+ β + γ = 2.1; α+ 0.7β + γ = 2.1; α+ β + 0.7γ = 2.1;13α = 7; 13β = 7; 13γ = 7;α = 1; β = 1; γ = 1.The proof of this theorem comes at the end of §6, and depends on work done in §4 and §5.Note that when projected onto one dimension, this theorem gives us a lower bound for S′substantially inferior to the one found in [1]. Specifically the above volume implies [1/3, 3/7] ⊆ S′.However, it does show that Q′ contains a volume, while the techniques used in [1] and [3] couldnot.4Chapter 3The Study of the Limit Points of Qwith Polynomials and BinaryFormsIn this chapter, we shall apply the techniques from [1] to the problem of finding the limit points ofQ′. The main theorem which the authors use to obtain most of their results is as follows:BFGS Theorem (Theorem 2 of [1]). 1. Let 1 ≤ Y ≤ X, where X is sufficiently large, and letf(x, y) ∈ Z[x, y] be a binary form whose irreducible factors fi are distinct and all have degreesnot exceeding µ. Let D denote the largest fixed divisor of f(x, y), and let S = D/Rad(D).Let N(X,Y ) denote the number of pairs (x, y) such thatX ≤ x ≤ 2X; Y ≤ y ≤ 2Yfor which f(x, y)/S is squarefree.Suppose for some  > 0, Xµ < (XY )3−, Y > X.ThenN(X,Y ) = CfXY{1 +O(1logX)}(3.1)where the constant Cf > 0 depends only on f , and the O constant depends only on .2. Suppose X = Y α, where α > 1 is fixed. It follows that 3.1 holds:(a) for any α when µ ≤ 3,5(b) if α < 3 when µ = 4,(c) if α < 3/2 when µ = 5.As described earlier, the BFGS Theorem is used in [1] to prove their main result, that uncondi-tionally, [1/3, 15/16] ⊆ S′.Ideally, one would hope that just as the BFGS Theorem is used to fill in most of the possible spaceof S′, the BFGS Theorem would do the same for Q′; but as the following proofs shall demonstrate,this is not the case. In particular, we shall find that their results naturally produce parameterized,one-dimensional curves in three dimensional space.3.1 Translation of the Results of [1] to Three DimensionsFirst, we shall apply the methods of [1] to find points in Q′.The following lemma uses the polynomial identities that gave [1] the result that [ 13 ,67 ] ⊆ S′, togive us the equivalent limit points of Q′:Lemma 3.1. For n = 1, 2, 4, 6, the line segments with endpoints(1n, 1,1n)→(0, 11n),lie in Q′.Proof. Let n ∈ {1, 2, 4, 6}.Observe that xn − yn has irreducible factors of degree at most 2. Thus by the BFGS Theoremfor any α > 1, there exist squarefree integer values of xy(xn − yn) for sufficiently large X,Y , withX = Y α, and X ≤ x ≤ 2X, Y ≤ y ≤ 2Y .Thus there exists a function f(Y, α) = (a, b) that maps each real Y (sufficiently large) and α > 0to two numbers a, b such thata = yn; b = xn − yn; a+ b = xn;Y α ≤ x ≤ 2Y α;Y ≤ y ≤ 2Y ;xy(xn − yn) is squarefree.6Consequently, it follows for Y sufficiently large that−→iq(f(Y, α)) =1n log x(log Rad y, log Rad(xn − yn), log Radx)=1n log x(log y, log(xn − yn), log x)=(log yn log x,log(xn − yn)n log x,1n)=(log Y +O(1)nα log Y +O(1),nα log Y +O(1)nα log Y +O(1),1n)=(1 +O(log−1 Y )nα+O(log−1 Y ),1 +O(log−1 Y )1 +O(log−1 Y ),1n).Thus it follows thatlimY→∞−→iq(f(Y, α)) =(1nα, 1,1n).In Q′ a line segment is obtained by considering all α > 1. By varying α, it is clear that it hasendpoints at(1n , 1,1n)and(0, 1, 1n).Hence, for n = 1, 2, 4, 6 the technique produces four line segments:(1, 1, 1)→ (0, 1, 1),(12, 1,12)→(0, 1,12),(14, 1,14)→(0, 1,14),(16, 1,16)→(0, 1,16).As can be seen, the techniques used in [1] cover most of S′, but only covers a set of four disjointline segments in the three dimensional case of Q′.We now move on to the binomial form that was used in [1] to show that [ 67 ,1213 ] ⊆ S′.Lemma 3.2. The line segment with endpoints (3/4, 1/2, 1/3) and (1/4, 1/2, 1/3) lies in Q′.Proof. As in [1], we start with the polynomial identityy3(2x+ y) + (x+ y)3(x− y) = x3(x+ 2y),and observe by cubing the variables thaty9(2x3 + y3) + (x3 + y3)3(x3 − y3) = x9(x3 + 2y3).Thus for f(x, y) = xy(x3 + 2y3)(2x3 + y3)(x3 + y3)(x3 − y3), we can apply the BFGS Theoremwith S = 2, µ = 3. Thus for all α > 1, and Y sufficiently large, given X = Y α there exist x, y ∈ Nsuch that X ≤ x ≤ 2X, Y ≤ y ≤ 2Y , and f(x, y)/2 is squarefree.Thus there exists a function g(Y, α) = (a, b) such thata = y9(2x3 + y3); b = (x3 + y3)3(x3 − y3);Y α ≤ x ≤ 2Y α, Y ≤ y ≤ 2Y, a, b are squarefree integers.7Observe that−→iq(g(Y, α)) =112α log Y +O(1)· (log Rad y9(2x3 + y3), log Rad(x3 + y3)3(x3 − y3), log Radx9(x3 + 2y3))=112α log Y +O(1)· (log y(2x3 + y3) +O(1), log(x3 + y3)(x3 − y3) +O(1), log x(x3 + 2y3) +O(1))=112α log Y +O(1)((3α+ 1) log Y +O(1), 6α log Y +O(1), 4α log Y +O(1))=((3α+ 1) log Y +O(1)12α log Y +O(1)6α log Y +O(1)12α log Y +O(1),4α log Y +O(1)12α log Y +O(1))=(1 +O(log−1 Y )4 +O(log−1 Y )+1 +O(log−1 Y )12α+O(log−1 Y ),1 +O(log−1 Y )2 +O(log−1 Y ),1 +O(log−1 Y )3 +O(log−1 Y )).Thus it follows, for all α > 1, thatlimY→∞−→iq(g(Y, α)) =(14+12α,12,13).Hence we obtain the endpoints (34,12,13),(14,12,13)in Q′.Finally, we work on extending the application of the polynomial identity and binary form thatwas used in [1] to show that [ 1213 ,1516 ] ⊆ S′ to the three dimensional problem.Lemma 3.3. The line segment with endpoints (2/5, 7/20, 7/20) and (2/5, 19/60, 7/20) lies in Q′.Proof. As described in [1], one has that(x+ y)7(x− y)(x2 − xy + y2) + y7(2x+ y)(3x2 + 3xy + y2) = x7(x+ 2y)(x2 + 3xy + 3y2).We shall note as an aside that [1] briefly discusses the origin of this identity, with further referencesfor those interested.Thus it follows that(x2 + y2)7(x2 − y2)(x4 − x2y2 + y4) + y14(2x2 + y2)(3x4 + 3x2y2 + y4)= x14(x2 + 2y2)(x4 + 3x2y2 + 3y4).8Observe that givenf1(x, y) = xy(x+ y)(x− y)(x2 + y2)(2x2 + y2)(x2 + 2y2),f2(x, y) = (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4),f(x, y) = f1(x, y)f2(x, y),as in [1], f(x, y) is separable, S = 6, µ = 4, and thus by the BFGS theorem, for all α ∈ (1, 3) andY sufficiently large with X = Y α, there exists x, y ∈ N such that X ≤ x ≤ 2X, Y ≤ y ≤ 2Y wheref(x, y)/6 is squarefree.Given k = gcd(a, b) ∈ [1, 6], there exists a function g(Y, α) = (a, b)/k such that for Y sufficientlylarge,a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)/k; b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)/k;Y α ≤ x ≤ 2Y α; Y ≤ y ≤ 2Y ; a, b are squarefree natural numbers.Thus one has that−→iq(g(Y, α)) =1log(a+ b)(log Rad a, log Rad b, log Rad a+ b)=120α log Y +O(1)(log((x2 + y2)(x2 − y2)(x4 − x2y2 + y4))+O(1),log(y(2x2 + y2)(3x4 + 3x2y2 + y4))+O(1),log(x(x2 + 2y2)(x4 + 3x2y2 + 3y4))+O(1))=(8α log Y +O(1)20α log Y +O(1),(6α+ 1) log Y +O(1)20α log Y +O(1),7α log Y +O(1)20α log Y +O(1)),=(2 +O(log−1 Y )5 +O(log−1 Y ),(3 + (2α)−1) +O(log−1 Y )10 +O(log−1 Y ),7 +O(log−1 Y )20 +O(log−1 Y )).Thus one has, for α ∈ (1, 3), thatlimY→∞−→iq(g(Y, α)) =(25,310+120α,720).Considering all possible values of α and noting that Q′ is a closed set, one has that this impliesthe existence of the line segment with endpoints(25,720,720),(25,1960,720)in Q′.9The paper [1] demonstrated that if the abc conjecture holds, S′ = [1/3, 1]. We shall now applythe same technique to determine what this implies for Q′. This technique does not rely on theBFGS Theorem, and is derived from the treatment of theorem 4 in [1].Lemma 3.4. If the abc conjecture holds, then for each n ∈ N, and t ∈ [0, 1] at least one of thefollowing points lies in Q′: (tn,1n, 1),(tn, 1,1n)Additionally, (0, 0, 1), (0, 1, 0) lie in Q′.Proof. Let n ∈ N, and letS1(Y, α) = {(x, y) ∈ N2 : Y α < x ≤ 2Y α, Y < y ≤ 2Y, xy(xn + yn) is squarefree},S2(Y, α) = {(x, y) ∈ N2 : Y α < x ≤ 2Y α, Y < y ≤ 2Y, xy(xn − yn) is squarefree},for α > 0.If one assumes the abc conjecture, it follows by the argument from the treatment of theorem 4found in [1] that for any Y sufficiently large and α > 1,S1(Y, α) 6= ∅ or S2(Y, α) 6= ∅.Thus it is the case that either there are infinitely many Y such that S1(Y, α) 6= ∅ or infinitelymany Y such that S2(Y, α) 6= ∅.1. Fix α, and suppose there exist infinitely many Y such that S1(Y, α) 6= ∅.It follows there exists a function F : (Y, α) 7→ (a, b), such that for Y sufficiently large, α > 1,one has thata = yn, b = xn, for some (x, y) ∈ S1(Y, α) if S1(Y, α) 6= ∅,or, to deal with the possibility that S1(Y, α) is sometimes empty,(a, b) = max{(m,n) ∈ N2 : (m,n) = F (Y ′, α) for some Y ′ ≤ Y, S1(Y, α) 6= ∅}.Observe that when S1(Y, α) 6= ∅,−→iq(F (Y, α)) =1log(xn + yn)(log Rad yn, log Radxn, log Rad(xn + yn))=1nα log Y +O(1)(log Y +O(1), α log Y +O(1), nα log Y +O(1))=(1 +O(log−1 Y )nα+O(log−1 Y ),1 +O(log−1 Y )n+O(log−1 Y ),1 +O(log−1 Y )1 +O(log−1 Y )),and that when S1(Y, α) = ∅, the function F remains constant.10Thus it follows thatlimY→∞−→iq(F (Y, α)) =(1nα,1n, 1).Thus the point ( 1nα ,1n , 1) lies in Q′.2. Fix α and suppose there exist infinitely many Y such that S2(Y, α) 6= ∅.It follows there exists a function F (Y, α) → (a, b) such that for Y sufficiently large, α > 1,one has thata = yn, b = xn − yn, for some (x, y) ∈ S2(Y, α) if S2(Y, α) 6= ∅,and behaves analogously to the function in the previous case when when S2(Y, α) = ∅. Observethat when S2(Y, α) 6= ∅,−→iq(F (Y, α)) =1log xn(log Rad yn, log Radxn − yn, log Radxn)=1nα log Y +O(1)(log Y +O(1), nα log Y +O(1), α log Y +O(1))=(1 +O(log−1 Y )nα+O(log−1 Y ),1 +O(log−1 Y )1 +O(log−1 Y ),1 +O(log−1 Y )n+O(log−1 Y )),and that when S2(Y, α) = ∅, the function remains constant. Thus it follows thatlimY→∞−→iq(F (Y, α)) =(1nα, 1,1n).Thus the point ( 1nα ,1n , 1) lies in Q′.This completes the proof of the first statement. For the second, simply observe thatlimn→∞−→iq(2n, 3n) = (0, 0, α)with α ≤ 1. But by the abc conjecture, 0 + 0 + α ≥ 1, so α = 1.Likewise,limn→∞−→iq(2n, 3n − 2n) = (0, α, 0)with α ≤ 1 and thus by the abc conjecture, α = 1.From this, it is clear that the results of [1] are not as capable of characterizing the set of limitpoints in the three-dimensional case as they were in the one-dimensional case. Their result that[1/3, 15/16] ⊆ S′ covers almost all of expected S′, but in the three dimensional case amounts to aset of disjointed line segments, covering no volume at all.11The same situation occurs when assuming the abc conjecture. In [1], it is shown that the abcconjecture implies [1/3, 1] = S′. The three dimensional equivalent of that statement would be thatthe abc conjecture implies that Q′ = H, but instead their technique only yields here that there existpoints split across pairs of line segments in Q′ that get arbitrarily close to the face of H nearest theorigin.The primary limitation of using the BFGS Theorem (and the abc variant) is that it naturallyproduces line segments. In the abc case, it was possible to produce an infinite number of suchsegments, giving us additional limit points that lie on the boundary of the heptahedron H, but stillno volume.However, it may be possible to extend this proof to either planes or volumes, but it will likelyrequire either modification of the theorem (for example, adding a third variable like Z ≤ z ≤ 2Z,Z = Xβ), or a method of adjusting the line segments by an  to “fill in” volumes of the heptahedron.Alternative binary forms or alternative factorizations of binary forms may be necessary.3.2 Applying the BFGS Theorem to Other Binary FormsWe shall now move on to look at some other polynomial identities and the limit points they producewhen the BFGS Theorem is applied. This section will show that a great diversity of lines and pointscan be found, but that this gets one no closer to showing a set with a positive volume lies in Q′.Lemma 3.5. The line segments (1, 12 ,12 )→ ( 12 , 12 , 12 ) and ( 34 , 12 , 14 )→ ( 12 , 12 , 14 ) both lie in Q′.Proof. Consider, for n = 1, 2, 3 the following binary formf(x, y) = xy(xn − yn)(2xn − yn).By taking f(x, y) and applying the BFGS Theorem, one observes that µ ≤ 3, and thus it followsthat for any sufficiently large Y , X = Y α, α > 1, there exist (x, y) such that X < x ≤ 2X,Y < y ≤2Y and f(x, y) is squarefree.Thus there exists a function F (Y, α) = (a, b) such thata = yn(2xn − yn); b = (xn − yn)2; a+ b = x2n; Y α < x ≤ 2Y α; Y < y ≤ 2Y ; f(x, y) is squarefree.12Thus one has that−→iq(F (Y, α)) =12n log x(log Rad yn + log Rad(2xn − yn), log Rad(xn − yn)2, log Radx2n)=12n log x(log y + log(2xn − yn), log(xn − yn), log x)=((1 + nα) log Y +O(1)2nα log Y +O(1),nα log Y +O(1)2nα log Y +O(1),α log Y +O(1)2nα log Y +O(1))=((1 + nα) +O(log−1 Y )2nα+O(log−1 Y ),1 +O(log−1 Y )2 +O(log−1 Y ),1 +O(log−1 Y )2n+O(log−1 Y )).Thus one has for any α > 1 that,limY→∞F (Y, α) =(12nα+12,12,12n).This gives the line segments: (1,12,12)→(12,12,12)(34,12,14)→(12,12,14)(46,12,16)→(12,12,16)If the above lemma could be proven to hold for all n ∈ N, one would have a sequence ofline segments that would eventually converge to the point (1/2, 1/2, 0), which lies on the criticalboundary. We shall note that it seems plausible in light of Lemma 3.4 that a proof that assumesthe abc conjecture might be capable of proving the above statement for all n ∈ N.Additional polynomial and binary form identities are likely to result is similar sorts of patterns,but it is not clear at all that these patterns can generalize sufficiently well so as to form volumes.3.3 Examining Q′ ∩ ([0, 1]3 \ H)From [3] one has the result that S′ ∩ [1, 3/2) 6= ∅.The analogous property for Q′ would be to show that for every point (x, y, z) where x+y+z = 1,the line segment (x, y, z)→ 2/3(x, y, z) is non-empty.First, we construct the three dimensional analogue of the argument given in [3] that shows thatS ∩ [1,∞) 6= ∅, for the purpose of illustrating the principle behind this technique.Lemma 3.6. Either there exist infinitely many points in the set {(0, x, 1n ) : n ∈ N, x ∈ [0, 1− 1n )}∩Q′, or else (0, 1, 0) ∈ Q′.13Proof. Fix some k ∈ N. For any squarefree n, letan = 1; bn = nk − 1; cn = nk.Observe that for any squarefree n, by considering all possible values of Rad(nk − 1), one hasthat−→iq(an, bn) =1k log n(log Rad 1, log Rad(nk − 1), log Radnk)=1k log n(0, log Rad(nk − 1), log n)=(0,log Rad(nk − 1)k log n,1k)∈ {(0, x, 1k) ∈ R3 : x ∈ [0, 1)}.By taking the subsequential limit of squarefree n, we observelimn→∞µ2(n)=1−→iq(an, bn) ∈ {(0, x, 1k) : x ∈ [0, 1]}.Thus it follows that for every k ∈ N, there exists a point of the form (0, x, 1k ) ∈ Q′ wherex ∈ [0, 1].If there are not infinitely many points in the set {(0, x, 1n ) : n ∈ N, x ∈ [0, 1− 1n )}, then it followsby the above that there are infinitely many points in the set {(0, x, 1n ) : n ∈ N, x ∈ [1− 1n , 1]}.Thus it follows by the fact that Q′ is closed that (0, 1, 0) ∈ Q′. This completes the proof.Next, we move on to the translation of the proof that S ∩ [1, 2] 6= ∅ from [3].Lemma 3.7. The set {(0, x, y) ∈ R3 : 12 ≤ x+ y ≤ 1} ∩Q′ is non-empty.Proof. For each n ∈ N, let tn be the smallest integer such thattn > 2n,Rad tn(tn − 1) ≤ 2tn,Rad tn(tn + 1) ≥ 2(tn + 1).By [3] tn exists for every n, and thus2(tn + 1) ≤ Rad(tn(t2n − 1)) ≤ 2tn(tn + 1),and thereforelog(tn) +O(1) ≤ log Rad(tn(t2n − 1)) ≤ 2 log tn +O(1).Thus one has that−→iq(1, (t2n − 1)) =12 log tn(0, log Rad(t2n − 1), log Rad(t2n)),14and finally,limn→∞−→iq(1, (t2n − 1)) ∈ {(0, x, y) ∈ R3 :12≤ x+ y ≤ 1}.Finally, we translate from [3] the proof that S′ ∩ [1, 3/2] 6= ∅.Lemma 3.8. The set {(x, y, z) ∈ R : 2/3 ≤ x+ y + z ≤ 1} ∩Q′ is non-empty..Proof. As described in [3], there are infinitely many x ∈ N such that f(x) = x(x − 1)(x − 3) issquarefree. Let t ∈ N,  > 0 be numbers such that f(t) is squarefree, t sufficiently large, and sufficiently small so it follows that Rad t(t−1)(t−3) ≥ t2+2. As described in [3], for each t,  thereexists a minimal integer m ∈ N such that(3m−1t)2+2 ≤ Rad 3mt(3mt− 1)(3mt− 3)(3mt− 9) ≤ (3mt)3+2.Thus for a = (3m−1t)2(3m−1t− 3), b = 3mt− 1, a+ b = (3m−1t− 1)3, one has that−→iq(a, b) =13 log(3mt− 1)(log Rad(3m−1t)2(3m−1t− 3), log Rad(3mt− 1), log Rad(3m−1t− 1)3)=13 log(3mt− 1)(log Rad(3m−1t)(3m−1t− 3), log Rad(3mt− 1), log Rad(3m−1t− 1)).As there are infinitely many such t, considering a, b as functions of t one has thatlimt→∞−→iq(a, b) ∈ {(x, y, z) ∈ R : 2/3 + 2/3 ≤ x+ y + z ≤ 1 + 2/3}.As this holds for  > 0 arbitrarily small, one has that there exists a point in{(x, y, z) ∈ R : 2/3 ≤ x+ y + z ≤ 1} ∩Q′ 6= ∅.The structure of this proof is such that it gives no information in Q′ that cannot be gleanedfrom the existence of a limit point in S′.So, a direct translation of the technique from S′ to Q′ does not work, but perhaps a variantthat chooses inequalities pairwise would. Given some of the flexibility that is afforded by the here, it may be useful to have two or more small variables to cover the volume contained inx = 0, y = 0, z = 0, x+ y + z = 1.153.4 Limitations of Current Polynomial MethodsThe subject of this work is to find and state the volume of Q′. As can be seen earlier, the methodsderived from [1] and [3] produce only line segments, and extending those segments to volumes doesnot seem to be possible without refinement of those techniques, or the determination of an infiniteset of binomial forms compatible with the BFGS Theorem whose endpoints are dense in somesurface contained in H.Despite some investigation, it is not clear that such a set of binomial forms is known to exist.Some general sets of binomial forms used in [1] are not of this form in general, even if some chosenexamples satisfy the requirements of the BFGS Theorem.We shall now move on to discussing potential avenues for improvements to the BFGS Theoremthat are plausible.First, to discuss one such improvement, we need to define the following:Definition 3.9 (Local Quality). Let {an}, {bn} ⊆ N be sequences of natural numbers.The set of local qualities, Lq of {(an, bn)} is defined to beLq({an, bn}) ={log Rad anlog bn: n ∈ N}′,that is to say, the set of limit points of real values of the formlog Rad anlog bn.As a matter of notation, Lq({an}) = Lq({an, an}).It is obvious that if {bn} is a subsequence of {an}, Lq({bn}) ⊆ Lq({an}).One improvements to the BFGS Theorem could involve extending it to demonstrate the existenceof xn, yn such that lim supLq({f(xn, yn)}) is equal to a fixed number β. This kind of extensionwould require some caution in its formulation as additional restrictions on the binary form and thevalue β would be needed to prevent contradicting the abc conjecture.Some attempts to do this by way of a scaling sequence an ultimately failed due to the difficultyin finding a generalizable polynomial identity that could incorporate an and produce suitable abctriples.Another possibility is to extend the BFGS Theorem to forms with three or more variables.Extending this result to 3 or 4 variable problems is difficult, and approaches that attempt to makef into a variable run into problems ensuring that the variables used to alter f remain of relevantscale as (x, y) become arbitrarily large. For example, one natural extension of f(x, y) uses u, v, β, γas variables (u, v squarefree):f(x, y) = (xuyv((xu)n − (yv)n).16It does not seem possible to naturally extend the BFGS Theorem by relating uβ , vγ to the sizeof x, y in any way: the structure of the proofs found in [1] are of no help, as they rely on startingwith arbitrarily large x, y for each given f . Furthermore, this function would only produce a planeat best, despite the addition of two variables.However, if for some α > 1 one could show that for any β ∈ [0, 1] given that there exist infinitelymany square-free (x, y) where x ≈ yα that there exists a subsequence where the ratio log Rad f(x, y)over log f(x, y) approaches some constant β ∈ [0, 1], this would then add one degree of freedom,giving for a = xn − yn, b = yn, c = xn:−→iq(a, b) = (β,1nα,1n).However, as discussed in the subsection below, there would be a significant constraint on the α(unless abc is false), for which this could be true, so any proof would need to take into account thatα ≤ (n+ 1)/n, meaning that the planes generated would be particularly useless for large n.Overall, this suggests that different methods need to be explored, and that is why the primaryfocus for the future will be on the circle method.3.5 Related Limit Points and Proof LimitationsIn this section, we take a more general look at how the problem of finding points in Q′ relates tothe problem of finding similar points in other sets.We first observe a subtle implication of the abc conjecture, related to these local qualities.Lemma 3.10. Assume the abc conjecture holds.Let α > 1, Rad(bn)α  an, and {an} be is an increasing sequence, and (an, bn) = 1 for alln ∈ N. For any x ∈ Lq(an, an + bn), y ∈ Lq(an + bn), one has thatx+ y ≥ 1− α−1.Proof. Simply observe thatlim infn→∞ iq(an, bn)= lim infn→∞log Rad anlog(an + bn)+log Rad bnlog(an + bn)+log Rad(an + bn)log(an + bn)= inf Lq(an, an + bn) + inf Lq(bn, an + bn) + inf Lq(an + bn).As Rad(bn)α  an, we have 0 ≤ inf Lq(bn, an + bn) ≤ α−1. By the abc conjecture, one has thatiq(an, bn) ≥ 1. Thus inf Lq(an, an + bn) + inf Lq(an + bn) ≥ 1− inf Lq(bn, an + bn), and thusinf Lq(an, an + bn) + inf Lq(an + bn) ≥ 1− α−1.17The lemma follows immediately from the definition of the infimums.Currently, every application of the BFGS Theorem involves choosing abc triples where b ≈ aαfor some α > 1, with the triple becoming smaller as α increases. This does have implications for theusefulness of proof techniques for finding limit points in Q. Since we believe that the abc conjectureis true, we should also believe that any proof technique that fails if the abc conjecture is true mustfail in practice.Any technique that produces sets of pairs (a, b) where bα  a, Lemma 3.10 ensures that forany limit point (x, y, z) ∈ Q′, x+ z > 1− 1/α, as a consequence of the abc conjecture. Thus if onewants to find limit points where x + z = 2/3, one must ensure α ≤ 3. This is consistent with thelines generated so far by these techniques, and shows that na¨ıve attempts to extend lines on theinwards is likely to fail if α can get arbitrarily large.18Chapter 4Technical ResultsThis chapter deals with several technical lemmas needed to obtain results in later sections. As theseresults are technical, and difficult to give motivation for, readers interested in the bigger picturemay wish to skip this chapter and reference it as needed.4.1 Sets of abc and almost abc triplesThe primary purpose of this section is to produce functions to compensate for the limitations of thecircle method, which cannot distinguish between triples that are relatively prime and those thatare not, and furthermore cannot restrict the components of the triple to being approximately thesame size.We shall define the notion of a pre-abc triple and a δ-triple and prove some of their properties.Definition 4.1. A pre-abc triple is a triple (a, b, c) such that a+ b = c, a, b, c ∈ N.A proper set of pre-abc triples is a set S of pre-abc triples where (a, b, c) ∈ S implies for d = (a, b)that (a/d, b/d, c/d) ∈ S.In the context of the circle method, we will be counting squarefree numbers that are not neces-sarily relatively prime, that is to say, a set of pre-abc triples.Definition 4.2. For a fixed δ > 0, a δ-triple is a pre-abc triple (a, b, c) ∈ N3 satisfying the followingcondition:There exists some M ∈ R such that M < c ≤ 2M and either a ≤M1−δ or b ≤M1−δ.This criterion will be used to ensure the triples counted are of appropriate size.First, we show that for any function that counts pre-abc triples, there exists a related functionthat counts those triples where the gcd is large.19Lemma 4.3. Let S be a set of pre-abc triples.Let S(x) = {(a, b, c) ∈ S : c ≤ x}, and let S(x) = {(a, b, c) ∈ S : gcd(a, b) > x, c ≤ x}.Let F (x) = #S(x), and let F(x) = #S(x).1. If S is proper either F (x) x or there exists an infinite T ⊂ S where T is a set of abc triples.2. For all x > 0, F(x) x2−.Proof. 1. Suppose there does not exist an infinite T ⊂ S where T is a set of abc triples. Hencethere are finitely many abc triples in S. Let N = max{c : (a, b, c) ∈ S, (a, b) = 1}. Hence,F (N) is greater than or equal to the number of all abc triples in S. Since for all x ∈ R, S(x)is proper, all pre-abc triples that are not abc triples are multiples of some abc triple in S(N).Hence F (x) = #S(x) ≤ F (N) · x x. Hence F (x) x.2. Observe that for fixed d there exist at most x/d multiples of d less than or equal to x.Hence for S(x, d) = {(a, b, c) ∈ S : gcd(a, b) = d, c ≤ x} we have |S(x, d)| ≤ x2/d2. AsS(x) =⋃d>x S(x, d),F(x) = #S(x) =∑d>xx2/d2 ≤ x2∫ ∞x−1y−2dy  x2x−.This completes the proof.Next, we consider functions that count the cardinality of subsets where one part of the triple issmall.Lemma 4.4. Let S be a set of pre-abc triples.Let δ > 0. Let F (x) = #{(a, b, c) ∈ S : a, b, c ≤ x}. Let Fδ(x) = #{(a, b, c) ∈ S : a, b, c ≤x, (a, b, c) is a δ-triple}.1. If S is proper then the set of δ-triples in S is proper.2. If x ≥ 21/δ+1, then Fδ(2x)− Fδ(x) ≤ 8x2−δ.Proof. To prove the first point, observe that if (a, b, c) is a δ triple, then for d = (a, b), a′ = a/d, b′ =b/d, c′ = c/d, since there exists M such that (without loss of generality) a ≤ M1−δ and M < c, itfollows a′ ≤M1−δ/d ≤ (M/d)1−δ, and M/d ≤ c′, so (a′, b′, c′) is a δ triple.Now, we prove the second point.Let S?(x) ⊆ S be defined to be the set {(a, b, c) ∈ S|x < c ≤ 2x}. Observe that F (2x) − F (x)counts exactly the triples (a, b, c) in S such that x < c ≤ 2x and thus F (2x)− F (x) = |S?(x)|. LetSδ(x) be the set of δ-triples in S(x). It likewise follows that |S?δ (x)| = Fδ(2x)− Fδ(x).20Let x ≥ 21/δ+1. It follows that δ triples in Sδ(x) cannot satisfy a ≤M1−δ and b ≤M1−δ for allpossible M ∈ [c/2, c) since a+ b ≤ x. Without loss of generality, assume a < b. Then it follows thatfor each triple (a, b, c) ∈ Sδ(x) there exists some M such that M < c ≤ 2M and a < M1−δ. Hencea ≤ M1−δ < c1−δ ≤ (2x)1−δ. Hence for each (a, b, c) ∈ Sδ(x), a < (2x)1−δ. Thus 1 ≤ a < (2x)1−δand thus, after multiplying by 2 to get the case where b < a, we have:|Sδ(x)| ≤ 2(2x)1−δ(2x) = 2(2x)2−δ ≤ 8x2−δ.Hence |Sδ(x)|  x2−δ. This completes the proof.As should be clear, once one subtracts from the count of a set of pre-abc triples the upper boundsfor sets where the gcd is large and the sets where one of the terms of the triple is small, one is leftwith a lower bound on the number of triples that are approximately relatively prime and containterms of approximately the same size.4.2 The G(q) function and related functionsIn [2], part of their application of the circle method involves a function G(q), which is also relevantto our proofs. This section concerns some results on the behaviour of this function that were notcovered in the original paper.First, we shall give a definition of the function G [2]:Definition 4.5. The function G : N→ R is a multiplicative function, where for a prime power p`:G(p`) =− 1(p2−1) if 1 ≤ ` ≤ 2,0 otherwise.As it is relevant to future results, we shall also note that µ2 is the characteristic function of thesquarefree integers.The first result of this section is a generalization of Lemma 3.1 from [2] for square-free numbers.This proof uses a different method than [2] but gives the same bounds.Lemma 4.6. Let n,m ∈ N, m > n. Let R > 0.If m > n, ∑q≥Rqn|G(q)|m  Rn−m+1/2.If m ≤ n, ∑1≤q≤Rqn|G(q)|m  Rn−m+1/2.21Proof. Observe that if G(q) 6= 0 then q = ab2 where a, b are squarefree integers.If m > n, then∑q≥Rqn|G(q)|m ≤∞∑a=1∑b≥√R/aµ2(a)µ2(b)anb2n|G(a)|m|G(b2)|m≤∞∑a=1µ2(a)an|G(a)|m∑b≥√R/aµ2(b)b2n|G(b2)|m≤∞∑a=1an−2m∑b≥√R/ab2n−2m∞∑a=1an−2m(R/a)n−m+1/2 Rn−m+1/2∞∑a=1a−m−1/2  Rn−m+1/2.If m ≤ n, then∑1≤q≤Rqn|G(q)|m ≤∑1≤a≤R∑1≤b≤√R/aµ2(a)µ2(b)anb2n|G(a)|m|G(b2)|m≤∑1≤a≤Rµ2(a)an|G(a)|m∑1≤b≤√R/aµ2(b)b2n|G(b2)|m≤∑1≤a≤Ran−2m∑1≤b≤√R/ab2n−2m∑1≤a≤Ran−2m(R/a)n−m+1/2 Rn−m+1/2∑1≤a≤Ra−m−1/2  Rn−m+1/2.This completes the proof.Due to its significance in later proofs, where it will be a component of the coefficient comprisingthe asymptotic formulas for functions counting a superset containing abc-triples, we shall resolvethe infinite sum where n = 1, m = 3:Lemma 4.7. The sum ω =∑∞q=1 ϕ(q)G(q)3 converges and ω ≈ 0.872985953173535.Furthermore, for any positive R,∑q≤Rϕ(q)G(q)3 = ω +O(R−3/2).22Proof. By Lemma 4.6, with n = 1,m = 3, one has that the infinite sum converges and that for all > 0,∞∑q=1ϕ(q)|G(q)|3 =∑q≤Rϕ(q)|G(q)|3 +O(R−3/2+),so ω exists.Furthermore, we shall observe that∞∑q=1ϕ(q)G(q)3 =∏p(1 + ϕ(p)G(p)3 + ϕ(p2)G(p2)3)=∏p(1− p− 1(p2 − 1)3 −p(p− 1)(p2 − 1)3)=∏p(1− (p+ 1)(p− 1)(p2 − 1)3)=∏p(1− 1(p2 − 1)2)and that since ∑p∣∣∣∣log(1− 1(p2 − 1)2)∣∣∣∣ ≤ ∞∑n=2∣∣∣∣log(1− 1(n2 − 1)2)∣∣∣∣and ∣∣∣∣log(1− 1(n2 − 1)2)∣∣∣∣ 1n4it follows that∑p | log(1 − 1(p2−1)2 )| < ∞ and thus∏p(1 − 1(p2−1)2 ) > 0. Thus the Euler productalso converges, and must converge to the same quantity ω.Now, givenωR =∏p≤R(1− 1(p2 − 1)2),and the fact that for any positive R,∑p>R∣∣∣∣log(1− 1(p2 − 1))∣∣∣∣ ≤∑n>R∣∣∣∣log(1− 1(n2 − 1)2)∣∣∣∣≤∑n>R2n−4≤∫ ∞R2x−4dx≤ 2R−3/3,23it follows thatω =∞∑q=1ϕ(q)G(q)3=∏p(1− 1(p2 − 1)2)= ωR∏p>R(1− 1(p2 − 1)2)= ωR exp∑p>Rlog(1− 1(p2 − 1)2)∈ [ωR exp(−2R−3/3), ωR].Furthermore, since the exponential has slope less than one over negative values, it follows thatω ∈ [ωR − 2R−3/3, ωR].By computation of ω100000, one has thatω ∈ [0.8729859531735338, 0.8729859531735357].In §6 some additional proofs related to functions derived from the G function are provided.They are not listed here as those variants of the G function depend on certain parameters derivedfrom an independent variable only relevant to the problem described in that chapter.24Chapter 5The Trivial Point ProblemIn this chapter, we move on to using the Circle Method to prove the existence of a limit point at(1, 1, 1) ∈ Q′. This method involves calculating a function F (N) which counts the number of triplesa + b = c (not necessarily relatively prime) such that a, b, c are squarefree and c ≤ N . Then weshow that the rate of growth of F (N) exceeds the possible rate of growth expected if the numberof triples a+ b = c where a, b, c are square-free and relatively prime is finite, which is what provesthe existence of the limit point.We are doing this to showcase the methods that will be used in later sections to tackle a morecomplex, general problem. We use a variant of the method described in [2] to show that the trivialpoint is a limit point in Q′. In this case, F (N) =∫ 10f(x;N)f(x;N)f(−x;N) dx, where f is thesquarefree counting function described in [2]. That is to say,f(x;N) =∑1≤n≤Nµ2(n)e(nx),where µ2 is the characteristic function of the squarefree integers.Throughout this chapter, we shall use the same major M and minor m arcs as found in [2]. Forreference, these depend on two parameters Q and N , with 1 ≤ Q ≤ N1/2/2:M(q, k) = {x ∈ [0, 1) : |qx− k| ≤ Q/N},M(Q;N) =⋃1≤q≤Qq⋃k=0(q,k)=1M(q, k),m(Q;N) = {x ∈ [0, 1) : ∀q ∈ N, k ∈ Z, (k, q) = 1, |qx− k| ≤ Q/N implies q > Q}.We now establish some bounds on the integral over the minor arcs.25Lemma 5.1. For all  > 0,∫m|f(x;N)2f(−x;N)| dx ≤ supx∈m|f(x;N)|∫m|f(x;N)|2 dx N2+Q−3/2 +N1+Q.Proof. From [2, Theorem 2], for any  > 0,∫m|f(x;N)|2 dx N1+Q−1/2 +N Q2.From a paper by D.I. Tolev [5, Theorem 1] one has that supx∈m |f(x;N)|  N1+Q−1.Thus by Ho¨lder’s inequality with p = +∞ and q = 1 and the above observations, the desiredstatement naturally follows.In addition to theG function (first discussed in Definition 4.5) we shall use the following functionsfrom [2]:I(x) =N∑n=1e(nx),f?(x) =ζ(2)−1G(q)I(x− a/q) if x ∈M(q, a) ⊂M,0 otherwise,∆(x) = f(x)− f?(x).For those unfamiliar with the circle method, I(x) is the standard notation for the exponential sumin these problems, and f? is the approximation of f used in [2].From the definitions given above, we shall observe that:∫Mf(x)2f(−x) dx =∫M(∆(x) + f?(x))2(∆(−x) + f?(−x)) dx=∫M∆(x)2∆(−x) dx+∫M2∆(x)∆(−x)f?(x) dx+∫M∆(x)2f?(−x) dx+∫M∆(−x)f?(x)2 dx+∫M2∆(x)f?(x)f?(−x) dx+∫Mf?(x)2f?(−x) dx.Next, we place an upper bound on the growth of the difference between the integrals of thetriple f and triple f?.Lemma 5.2. For every  > 0,∫Mf(x)2f(−x) dx−∫Mf?(x)2f?(−x) dx max{N3/2+Q3/4, N1+Q2}.26Proof. By Lemmata 3.2 and 4.2 in [2], we have that for all  > 0,∫M|∆(x)|2 dx N Q2, and∫M|f?(x)∆(x)| dx N1/2+Q3/4for any Q ∈ N where Q ≤ 12√N .Now, observe that by definition, for any x ∈M(q, a) that|f?(x)| ≤ |ζ(2)−1G(q)I(x− a/q)| ≤ ζ(2)−1|I(x− a/q)|  N.Likewise,|f(x)| ≤N∑n=11 N, |∆(x)| ≤ |f(x)|+ |f?(x)|  N.Thus we have: ∣∣∣∣∫M∆(x)2∆(−x) dx∣∣∣∣ ≤ supx∈M|∆(x)|∫M|∆(x)2|dx N1+Q2, (5.1)∣∣∣∣∫M∆(x)∆(−x)f?(x) dx∣∣∣∣ ≤ supx∈M|f?(x)|∫M|∆(x)2|dx N1+Q2, (5.2)∣∣∣∣∫M∆(−x)f?(x)2 dx∣∣∣∣ ≤ supx∈M|f?(x)|∫M|∆(x)f?(x)|dx N3/2+Q3/4, (5.3)∣∣∣∣∫M∆(x)2f?(−x) dx∣∣∣∣ = ∣∣∣∣∫M∆(x)∆(−x)f?(x) dx∣∣∣∣ N1+Q2, (5.4)∣∣∣∣∫M∆(x)f?(x)f?(−x) dx∣∣∣∣ = ∣∣∣∣∫M∆(−x)f?(x)2 dx∣∣∣∣ N3/2+Q3/4. (5.5)This completes the proof.Now, it suffices to calculate∫Mf?(x)2f?(−x) dx.Lemma 5.3. The main term resolves as follows:∫Mf?(x)2f?(−x) dx = N(N − 1)2ζ(2)−3(Q∑q=1ϕ(q)G(q)3)+O(N2Q−3/2).Proof. To start, recall that as in [2] for x ∈M(q, a), f?(x) = ζ(2)−1G(q)I(x−a/q), where G(n), I(x)are as defined earlier.27As M =⋃Qq=1⋃0≤a≤q(a,q)=1M(a, q), it follows that:∫Mf?(x)2f?(−x) dx =Q∑q=1q∑a=0(a,q)=1(G(q)ζ(2))3 ∫M(q,a)I(x− a/q)2I(a/q − x) dx=Q∑q=1q∑a=1(a,q)=1(G(q)ζ(2))3 ∫ Q/qN−Q/qNI(x)2I(−x) dx=Q∑q=1q∑a=1(a,q)=1(G(q)ζ(2))3(∫ 1/2−1/2I(x)2I(−x) dx+O(∫ 1/2Q/qN‖x‖−3 dx))=Q∑q=1ϕ(q)G(q)3ζ(2)−3(∫ 1/2−1/2I(x)2I(−x) dx+O(∫ 1/2Q/qN‖x‖−3 dx)).Note that the transition from a = 0 to a = 1 is facilitated by the fact that for q = 1 both intervalsare of half length, and since the exponential sum I(x) is periodic modulo Z, they can be stitchedtogether into a single full-length interval. We now compute the integral for the main term:∫ 1/2−1/2I(x)2I(−x) dx =∫ 1/2−1/2(N∑n=1e(nx))2( N∑m=1e(−mx))dx=∫ 10N∑m=1e(−mx)(N∑n=1e(nx))2dx=N∑m=1∫ 10(N∑n=1e(nx))2e(−mx) dx =N∑m=1(m− 1) = N(N − 1)2.Now, for the integral in the error term:∫ 1/2Q/qN‖x‖−3 dx q2N2Q−2.By Lemma 4.6 one has thatQ∑q=1ϕ(q)G(q)3∫ 1/2Q/qN‖x‖−3 dxQ∑q=1q|G(q)|3q2N2Q−2  N2Q−2Q∑q=1q3|G(q)|3  N2Q−3/2+.Hence, it follows that:∫Mf?(x)2f?(−x) dx = N(N − 1)2ζ(2)−3(Q∑q=1ϕ(q)G(q)3)+O(N2Q−3/2+).28Putting it all together, the following theorem results:Theorem 5.4. The following holds for any 1 < Q ≤ 12N1/2 and all  > 0:F (N) = ωζ(2)−3N(N + 1)/2 +O(N2+Q−3/2 +N3/2+Q3/4 +N1+Q2).Proof. By Lemmata 5.3 and 4.7, for all  > 0:∫Mf?(x)2f?(−x) dx = ωζ(2)−3N(N + 1)/2 +O(N2Q−3) +O(N2Q−3/2)Hence, we conclude that for all  > 0, that∫Mf?(x)2f?(−x) dx = ωζ(2)−3N(N + 1)/2 +O(N2Q−3/2).Combining this result with Lemmata 5.1 and 5.2 one has that:F (N) =∫ 10f(x)2f(−x) dx=∫Mf?(x)2f?(−x) dx+O(N3/2+Q3/4 +N1+Q+N2+Q−1/2)= ωζ(2)−3N(N + 1)/2 +O(N2Q−3/2 +N3/2+Q3/4 +N1+Q+N2+Q−3/2) +N1+Q2.Noting that N2Q−3/2  N2+Q−3/2 and N1+Q  N1+Q2 for any Q > 1 completes theproof.Corollary 5.5. For Q = N2/9, and N sufficiently large,F (N) = ωζ(2)−3N2/2 +O(N5/3+).This suffices to show that the equation a + b = c for a, b, c ∈ N has infinitely many square-freesolutions. However, we need to show that a + b = c for a, b, c ∈ N has infinitely many square-freesolutions where (a, b) are relatively prime.Theorem 5.6. There exist infinitely many abc triples where a, b, c are squarefree, and (1, 1, 1) ∈ Q′.Proof. Observe that by Corollary 5.5F (M) = ωζ(2)−3(M2 +M)/2 +O(M5/3+)F (2M) = ωζ(2)−3(2M2 +M) +O(M5/3+)DF (M) = F (2M)− F (M) = ωζ(2)−3(3M2 +M)/2 +O(M5/3+).Let AF (M, ) count the subset of squarefree pre-abc triples on the interval (M, 2M ] wheregcd(a, b) > M . Let BF (M, δ) count the subset of squarefree pre-abc δ-triples on the interval29(M, 2M ]. By Lemma 4.3, for any  > 0, AF (M, ) = O(M2−). By Lemma 4.4, for any δ > 0,BF (M, δ) = O(M2−δ).Now, fix δ > 0 and observe that D(M, δ) = DF (M) − AF (M, δ/2) − BF (M, δ/2) = DF (M) +O(M2−δ/2), which is strictly greater than zero for sufficiently large M .For M sufficiently large, D(M, δ) > 0 and thus there exists a triple (a, b, c) such that d =gcd(a, b) ≤Mδ/2, a, b ≥ cδ/2, and M < c ≤ 2M .Now, let (a′, b′, c′) = (a/d, b/d, c/d). It follows that a′ ≥ aM−δ/2, b′ ≥ bM−δ/2, c′ ≥ cM−δ/2 >M1−δ/2. Furthermore, a′ ≥ aM−δ/2 ≥ c1−δ/2M−δ/2 ≥ c1−δ ≥ (c′)1−δ. By symmetric argument,b′ ≥ (c′)1−δ.Since M can be made arbitrarily large, and c′ > M1−δ/2 it follows that there are infinitely manysquarefree relatively prime triples (a, b, c) such that a ≥ c1−δ, b ≥ c1−δ where a+ b = c. Note thatthese triples are such that−→iq(a, b) ∈ [1− δ, 1]2 × {1} ⊂ R3.Since δ can be made arbitrarily small, this suffices to prove the existence of the limit point(1, 1, 1) in Q.This result is the end goal of this chapter. We have used the circle method to show that thereexist infinitely many relatively prime positive squarefree integers a, b such that a+b is also squarefreeand of similar size.30Chapter 6Determination of a Volume in QIn this chapter, we shall use the same methods as in the Trivial Point Problem, but apply them toa more general problem, so as to allow the computation of a large volume of points in Q.This chapter concerns the analysis of a counting function for a set of natural number triplesa+ b = c which, in the limit as N →∞ have −→iq(a, b) = (α, β, γ). Before giving a precise definitionof our counting function, we will spend some time discussing the strategy of this proof.We will employ the circle method to compute an asymptotic formula for a function F (N) thatdepends on six parameters: α, β, γ and pa, pb, pc. The parameters α, β, γ specify the limit point inQ we seek to prove exists, and pa, pb, pc are arbitrary distinct prime numbers. We have allowedfor arbitrary distinct prime numbers instead of choosing specific primes (such as 2, 3, 5) as it addslittle complexity to the proof and greater generality.The asymptotic formula of our counting function F (N) will itself be insufficient to prove theexistence of a limit point. This is because our function will count triples (a, b, c) that are notnecessarily relatively prime to each other, and will count some numbers that are too small in acertain sense. However, these issues can be overcome, as the improper triples can be excluded bymeans of counting. Once this is done, constructing a limit point is simple.6.1 DefinitionsWe now move on to the formal definitions.Let α, β, γ ∈ (0, 1], and pa, pb, pc ∈ N be distinct prime numbers. The choice of primes isarbitrary. Our proofs will show that the choice of primes affects the constant coefficients of themain term, and little else.Now, we can move on to the definitions, with F (N), we shall first give a small list of quantities31that are functions of N and the parameters α, β, γ which will be used in the definition of F (N) andin later proofs.Definition 6.1 (Quantities that Depend on N). For any given N ∈ R≥1, letA = b(1− α) logpa Nc, B = b(1− β) logpb Nc, C = b(1− γ) logpc Nc.Let ha = N1−αp−Aa , hb = N1−βp−Bb , and hc = N1−γp−Cc .Note that ha, hb, hc are technically functions of N , and that ha : R>0 → [1, pa), hb : R>0 →[1, pb), and hc : R>0 → [1, pc). It should also be clear that they are periodic on a logarithmic scale.We now have enough to give a precise definition of F (N).Definition 6.2 (The Counting Function F (N)). The function F (N) counts all triples of the form(pAa a, pBb b, pCc c) where a ≤ Nα, b ≤ Nβ , c ≤ Nγ and a, b, c are squarefree. We shall also describeF (N) as an integral of the function f(x;M) as from [2]. The function F (N) is thus represented bythe following integral:F (N) =∫ 10f(pAa x;Nα)f(pBb x;Nβ)f(−pCc x;Nγ)dx.Note that F (N) depends only on the choice of (α, β, γ) ∈ [0, 1]3 and pa, pb, pc. Once these arefixed, A,B,C are simply functions of N . Likewise, an application of the circle method requires adissection of the unit interval [0, 1) into major arcs. For reasons that will become apparent, thedissection used will be more complex than is usual, with three major arc components. For thesemajor arcs, there will be three parameters Qa, Qb, Qc ∈ R determining their size such that1 ≤ Qa ≤ 12Nα/2, 1 ≤ Qb ≤ 12Nβ/2, 1 ≤ Qc ≤ 12Nγ/2.Definition 6.3 (Major Arcs). The major arcs are defined in terms of the following componentsets:Mα(q, a) = {x ∈ [0, 1) : |qpAa x− a| ≤ QaN−α},Mβ(q, b) = {x ∈ [0, 1) : |qpBb x− b| ≤ QbN−β},Mγ(q, c) = {x ∈ [0, 1) : |qpCc x− c| ≤ QcN−γ}.These components become the local major arcsMα =⋃q≤QapAa q⋃a=0(q,a)=1Mα(q, a), Mβ =⋃q≤QbpBb q⋃b=0(q,b)=1Mβ(q, b), Mγ =⋃q≤QcpCc q⋃c=0(q,c)=1Mγ(q, c).32The entire major arc is thus defined to be the intersection of the three local major arcs, M =Mα ∩Mβ ∩Mγ . Thus one can say that the major arcs constitute the union of all intersectionsof Mα(qa, a), Mβ(qb, b) and Mγ(qc, c) where qa ≤ Qa, qb ≤ Qb, qc ≤ Qc and (qa, a) = (qb, b) =(qc, c) = 1.Furthermore, an important subset of the major arcs, which shall be called the harmonic majorarcs MH , will have componentsMH(q, k) = Mα(q(q, pAa ),pAa(q, pAa )k)∩Mβ(q(q, pBb ),pBb(q, pBb )k)∩Mγ(q(q, pCc ),pCc(q, pCc )k).And finally,MH =⋃1≤qq/(q,pAa )≤Qaq/(q,pBb )≤Qbq/(q,pCc )≤Qcq⋃k=0(q,k)=1MH(q, k).Note that MH is clearly a subset of M by definition since it is a union of intersections ofMα(qa, a), Mβ(qb, b) and Mγ(qc, c) where by definition qa = q/(q, pAa ) ≤ Qa, qb = q/(q, pBb ) ≤Qb, qc = q/(q, pcc) ≤ Qc and (qa, a) = (qb, b) = (qc, c) = 1.The choice of the word harmonic to describe MH is due to the fact that, as will be proven later,the centres of the intervals are precisely the rational numbers such that one arc component of eachtype is centred on them, and this is considered a sort of “harmony.” The harmonic major arcs arethe arcs over which the asymptotic formula of F (N) will actually be calculated.As a shorthand, we will sometimes writeQ˜ = {q ∈ N : q/(q, pAa ) ≤ Qa, q/(q, pBb ) ≤ Qb, q/(q, pCc ) ≤ Qc}to denote the set of natural numbers such that MH =⋃q∈Q˜⋃qk=0(q,k)MH(q, k).Unlike other arc dissections, which split [0, 1) into the major arcs and minor arcs, we shall divide[0, 1) into four different categories, the major arcs denoted M as defined above, the semi-majorarcs denoted ∂M, the semi-minor arcs ∂m, and the minor arcs m. While it is entirely possible todefine these arcs as unions of intersections of the major arcs components, this somewhat obscurestheir structure. Instead we shall define them using the following characteristic functions:33Definition 6.4 (Characteristic Functions). Defineχα(x) =1 if x ∈Mα(qa, a) where (qa, a) = 1, 1 ≤ qa ≤ Qa,0 otherwise,χβ(x) =1 if x ∈Mβ(qb, b) where (qb, b) = 1, 1 ≤ qb ≤ Qb,0 otherwise,χγ(x) =1 if x ∈Mγ(qc, c) where (qc, c) = 1, 1 ≤ qc ≤ Qc,0 otherwise,χ(x) = χα(x) + χβ(x) + χγ(x).From the definition of the major arcs, it follows that x ∈ M if and only if χ(x) = 3. We thendefine the major, semi-major, semi-minor, and minor arcs as follows:Definition 6.5 (Other Arcs).M = {x[0, 1) : χ(x) = 3},∂M = {x ∈ [0, 1) : χ(x) = 2},∂m = {x ∈ [0, 1) : χ(x) = 1},m = {x ∈ [0, 1) : χ(x) = 0}.It should be obvious that these four sets are disjoint, and that M ∪ ∂M ∪ ∂m ∪m = [0, 1).We shall now define the harmonic semi-major arcs, as well as a split of them into a low andhigh component. These serve a role analogous to the harmonic majors.Definition 6.6 (Harmonic Semi-Major Arcs). Let∂Q˜ ={q ∈ N : at least two of the following hold: q(q, pAa )≤ Qa, q(q, pBb )≤ Qb, q(q, pCc )≤ Qc}.For any given q ∈ N, k ∈ Z define the harmonic semi-major component∂MH(q, k) = ∂M ∩(Mα(q(q, pAa ),pAa(q, pAa )k)∪Mβ(q(q, pBb ),pBb(q, pBb )k)∪Mγ(q(q, pCc ),pCc(q, pCc )k)).The harmonic semi-major arcs ∂MH are the arcs such that∂MH =⋃q∈∂Q˜q⋃k=1(q,k)=1∂MH(q, k).34Finally, we shall define the low harmonic semi-major arcs to be∂MH0 =⋃q∈Q˜q⋃k=1(q,k)=1∂MH(q, k),and also the high harmonic semi-major arcs to be∂MH1 = ∂MH \ ∂MH0.In addition to the arcs defined above, it is necessary to define the reduced major arcs and reducedminor arcs which are not subsets of the arcs listed above, but serve as a bridge to results from [2].In particular, the reduced major arcs are equivalent to the major arcs from [2] and the reducedminor arcs are equivalent to the minor arcs from [2].Definition 6.7 (Reduced Major Arcs). Let M ∈ R≥1, and Q ∈ R where 1 ≤ Q ≤ 12N1/20 .The reduced major arcs at Q;M are M?(Q;M) and are defined to be aM?(Q;M) =⋃q≤Qq⋃k=1(q,k)=1M?(q, k;Q;M),withM?(q, k;Q;M) = {x ∈ [0, 1) : |qx− k| ≤ QM−1}.Likewise, the reduced minor arcs at Q;M are m?(Q;M) = [0, 1) \M?(Q;M), or in other words,m?(Q;M) = {x ∈ [0, 1) : ∀q ∈ N, a ∈ Z, (q, a) = 1, if |qx− a| ≤ QM−1 then q > Q}.The reduced majors arcs and reduced minor arcs have definitions equivalent to the major arcsand minor arcs found in [2] but with different parameters, and on occasion some proofs will translatethe complex major arcs of this paper into the reduced major arcs to apply results from [2].We move on to defining some functions that will be used.Definition 6.8. DefineG?3(q) = G(q(q, pAa ))G(q(q, pBb ))G(q(q, pCc )),Iα(x) =∑n≤Nαe(nx), Iβ(x) =∑n≤Nβe(nx), Iγ(x) =∑n≤Nγe(nx).We shall then define the following functions, all analogous to the f? function in [2] over different35reduced major arcs.f?α(x) =ζ(2)−1G(q)Iα(x− k/q) if x ∈M?(q, k;Qa;Nα) ⊂M?(Qa;Nα),0 otherwise.f?β(x) =ζ(2)−1G(q)Iβ(x− k/q) if x ∈M?(q, k;Qb;Nβ) ⊂M?(Qb;Nβ),0 otherwise.f?γ (x) =ζ(2)−1G(q)Iγ(x− k/q) if x ∈M?(q, k;Qc;Nγ) ⊂M?(Qc;Nγ),0 otherwise.We shall also define the following functions, analogous to the ∆(x) functions in [2].∆α(x) = f(x;Nα)− f?α(x),∆β(x) = f(x;Nβ)− f?β(x),∆γ(x) = f(x;Nγ)− f?γ (x)Finally, unlike the trivial point problem discussed in the previous chapter, the central componentof the main term is not an easily computed integral. The main term will be evaluated first in termsof this integral, and then this integral will be evaluated separately.Definition 6.9 (Main Term Integral U(N)). DefineU(N) =∫ 10Iα(pAa x)Iβ(pBb x)Iγ(pCc x)dxWe also have a related function, given below.Definition 6.10. For every δ > 0, defineUδ(N) = #{(a, b, c) ∈ N : a ≤ Nα−δ, b ≤ Nβ−δ, c ≤ Nγ−δ, pAa + pBb = pCc c}.We also have another two functions related to F .Definition 6.11. We have for N2 being the set of squarefree natural numbers,F (N, d) = #{(a, b, c) ∈ N32 : pAa a+ pBb b = pCc c, gcd(a, b) = d, a ≤ Nα, b ≤ Nβ , c ≤ Nγ}.and the functionF(N) = #{(a, b, c) ∈ N32 : pAa a+ pBb b = pCc c, gcd(a, b) ≥ N }.36Note that f?α, f?β , f?γ functions are analogous to the f? function defined in [2], over differentreduced major arcs. Likewise the ∆α,∆β ,∆γ functions are analogous to the ∆ function definedin [2], over different reduced major arcs with the corresponding f? function.We shall end the section with an definition of certain useful conditions in the calculation ofthe major and semi-major arcs and their relationship with their harmonic components. The use ofthese definitions is not readily apparent or obvious, but later in this work we shall show that whenQa  Nλaα; Qb  Nλbβ ; Qc  Nλcγ ;the weaker condition implies that the major arcs are equal to the harmonic major arcs, and thestronger condition implies that the semi-major arcs are equal to the harmonic semi-major arcs. Itwill be shown later that both of these conditions must be satisfied to find the asymptotic formulaof the integral at the heart of the problem.Definition 6.12. LetDH(λa, λb, λc) = {(1− λa)α+ (1− λb)β, (1− λa)α+ (1− λc)γ, (1− λb)β + (1− λc)γ},be known as the harmonic lambda factor.If maxDH(λa, λb, λc) > 1, then DH(λa, λb, λc) is called weakly harmonic.If minDH(λa, λb, λc) > 1, then DH(λa, λb, λc) is called strongly harmonic.6.2 Evaluation of U(N), and Bounds For Uδ and FThe function U(N) shall be evaluated by a simple argument involving the Geometry of Numbers.We shall quote the following lemma proven in a paper by Kane [4], which deals with a similarproblem, and is perfectly suited to solving this one:Lemma 6.13 (Lemma 3 of [4]). Let L be a lattice in a two-dimensional vector space V , and P aconvex polygon in V . Let m be the minimum separation between points in L. Then#(L ∩ P ) = Volume(P )Covolume(L)+O(Perimeter(P )m+ 1).It should be noted that this lemma, or its proof by Kane, does not in any way depend on whichpoints in the boundary of P lie in P , as the number of points on the boundary is clearly containedin the error term.Lemma 6.14. Let VN = {(x, y, z) ∈ R3 : pAa x + pBb y = pCc z}. Note that the dependence on N isvia A,B,C.37Let ρ =√1 +h2ch2aN2α−2γ + h2ch2bN2β−2γ .For any polygon P ⊂ VN it follows that if P ′ is the projection of P onto the xy plane, thenVolume(P ) = ρVolume(P ′).Proof. Observe that for any (x, y, z) ∈ VN , by rearrangement one has that z is a function of x, yand thusz =pAapCcx+pBbpCcy,∂z∂x=pAapCc=hchaNα−γ ,∂z∂y=pAapCc=hchbNβ−γ .Hence it follows that if P ⊂ VN is a polygon and P ′ is the projection of R onto the xy-plane,Volume(P ) =∫∫P ′√1 +(∂z∂x)2+(∂z∂y)2dA =∫∫P ′ρ dA = ρVolume(P ′).Lemma 6.15. Let N > 0, PN ⊆ R3 be the polygon produced by the intersection of the solidrectangular prism RN = (0, Nα]×(0, Nβ ]×(0, Nγ ] and the plane VN = {(x, y, z) ∈ R3 : pAa x+pBb y =pCc z}.Let ρ =√1 +h2ch2aN2α−2γ + h2ch2bN2β−2γ .Then the polygon PN is convex and:1. If ha ≤ hc and hb ≤ hc then PN is a triangle andVolume(PN ) =12hahbh2cρNα+β ,Perimeter(PN ) =√h2ah2cN2α +N2γ +√h2ah2cN2α +h2bh2cN2β +√h2bh2cN2β +N2γ .2. If ha ≤ hc and hb > hc then PN is a quadrilateral andVolume(PN ) =(hahc− ha2hb)ρNα+β ,Perimeter(PN ) =√h2ah2cN2α +N2γ +√h2ah2bN2α +N2β+√(hahc− hahb)2N2α +(1− hchb)2N2γ +√N2β +h2ch2bN2γ .383. If ha > hc and hb ≤ hc then PN is a quadrilateral andVolume(PN ) =(hbhc− hb2ha)ρNα+β ,Perimeter(PN ) =√N2α +h2ch2aN2γ +√(hbhc− hbha)2N2β +(1− hcha)2N2γ+√N2α +h2bh2aN2β +√h2bh2cN2β +N2γ .4. If ha > hc and hb > hc and(hahc− hahb)≥ 1, then PN is a parallelogram andVolume(PN ) = ρNα+β ,Perimeter(PN ) = 2√N2α +h2ch2aN2γ + 2√N2β +h2ch2bN2γ .5. If ha > hc and hb > hc and(hahc− hahb)< 1 then PN is a pentagon andVolume(PN ) =(1− hb2ha(1− hahc+hahb)2)ρNα+β ,Perimeter(PN ) =√N2α +h2cha2N2γ +√(hbhc− hbha)2N2β +(1− hcha)2N2γ+√(1− hahc+hahb)2N2α +(1− hbhc+hbha)2N2β+√(hahc− hahb)2N2α +(1− hchb)2N2γ +√N2β +h2ch2bN2γ .Furthermore,12p2cNα+β ≤ ρ−1 Volume(PN ) ≤ Nα+β .Proof. First, note given that N > 0, RN is a three dimensional solid with one vertex at the originand one vertex (Nα, Nβ , Nγ) in the first octant. Since VN is a plane containing the origin with allnormal vectors pointing outside the first octant it follows that the intersection of RN and VN isnon-empty. Thus PN is a nonempty area lying in VN , with (0, 0, 0) on the boundary of PN .Since RN is a convex volume, and VN is a plane, and PN a non-empty intersection of the two,it follows that PN is a convex polygon whose edges lie on faces of RN and whose vertices lie on theedges of RN .39One may see that the five shapes listed in the lemma are exhaustive by the following argument.PN contains the origin and two line segments from (0, 0, 0) that extend along the x = 0 and y = 0faces of the rectangle. These segments will either have endpoints on the top or sides of the rectangle.If both endpoints are on the top, the resulting shape is a triangle. If one endpoint is on thetop, the resulting shape is a quadrilateral. If neither endpoint is on the top, then one either getsa parallelogram or a pentagon, depending upon whether or not the plane intersects the top face ofthe rectangle at all.The exact values given in the Lemma arise from the computation of the volume and perimeterof the polygons from the above description. For a full treatment of the geometric problem, seeAppendix A.Lemma 6.16. Let ΛN = {(x, y, z) ∈ Z : pAa x + pBb y = pCc z} be a lattice lying the plane VN ={(x, y, z) ∈ R : pAa x+ pBb y = pCc z}.Let ρ =√1 +h2ch2aN2α−2γ + h2ch2bN2β−2γ .It follows thatCovolume(ΛN ) =ρhcN1−γ ,and there exists some ΨN ∈ N such that ΨN < pCc , (ΨN , pc) = 1 and the vectors~vN1 =(1,ΨN ,pAapCc+pBbpCcΨN), ~vN2 =(0, pCc , pBb),form a basis for ΛN .Proof. Before beginning with the proof, note that when dealing with equivalence classes modulo anatural number, a negative integer exponent −n is understood to be the multiplicative inverse tothe power of n in the appropriate group of residue classes. Let ΨN ∈ N be the number such thatΨN ≡ −p−Bb pAa (mod pCc ) and ΨN < pCc .For any (a, b, c) ∈ ΛN it holds that pAa a+ pBb b ≡ 0 (mod pCc ), so it follows that since (pb, pc) =1, and b ≡ −p−Bb pAa a ≡ ΨNa (mod pCc ), we thus have for some k ∈ Z that b = ΨNa + kpCc .Consequently, since for any (a, b, c) ∈ ΛN ,c =pAapCca+pBbpCcb =pAapCca+pBbpCcΨNa+ pBb k.it follows that any (a, b, c) ∈ ΛN can be represented solely in terms of (a, k) and thusv1 =(1,ΨN ,pAapCc+pBbpCcΨN),v2 =(0, pCc , pBb).40forms a basis for ΛN .The projection of v1, v2 onto the xy plane produces a lattice with basis (1,ΨN ), (0, pCc ) whichsince 1 ≤ ΨN < pCc forms a parallelogram S of area pCc . By Lemma 6.14, it follows that the area ofthe parallelogram produced by v1, v2 is ρpCc . Thus, as the parallelogram produced by v1, v2 is thefundamental domain of ΛN ,Covolume(ΛN ) =ρhcN1−γ .Now, we have sufficient tools to determine the value of U(N), as defined in Definition 6.9. Thecalculation of this value is critical, as it is in fact going to be the main term of the asymptoticformula of F (N), our counting function.Lemma 6.17. For any real N > 0,U(N) = H(N)Nα+β+γ−1 +O(Nα +Nβ +Nγ),H(N) =hahb2hcif hb ≤ hc and ha ≤ hc,hb(1− hc2ha)if hb ≤ hc and ha > hc,ha(1− hc2hb)if hb > hc and ha ≤ hc,hc if hb > hc and ha > hc andhahc− hahb ≥ 1,hc(1− hb2ha(1− hahc + hahb)2)if hb > hc and ha > hc andhahc− hahb < 1.In particular,12pcNα+β+γ−1 ≤ U(N) +O(Nα +Nβ +Nγ) ≤ Nα+βγ−1.Proof. First, note by the definition of the integral, U(N) counts all triples (a, b, c) ∈ N3 such thatthat a ≤ Nα, b ≤ Nβ , c ≤ Nγ and pAa a+ pBb b = pCc c.Thus for the lattice ΛN = {(a, b, c) ∈ Z : pAa a + pBb b = pCc } lying in the plane VN = {(x, y, z) ∈R : pAa x + pBb y = pCc z}, and the polygon PN = VN ∩ (0, Nα] × (0, Nβ ] × (0, Nγ ] we have thatU(N) = #(PN ∩ ΛN ).Thus by Lemma 6.13 (Kane),U(N) =Volume(PN )Covolume(ΛN )+O(Perimeter(PN )m+ 1).As m is the minimum separation between the points of ΛN , and ΛN is a lattice with integercoefficients, 1/m ≤ 1. By Lemma 6.15, for every possible shape PN could take, Perimeter(PN ) Nα +Nβ +Nγ . Thus it follows thatU(N) =Volume(PN )Covolume(ΛN )+O(Nα +Nβ +Nγ).41To complete the proof of the formula for U(N), for each possible shape of PN given the param-eters ha, hb, hc, we divide the volume of PN from Lemma 6.15 by the covolume of ΛN from Lemma6.16 to get the equation given by the lemma. Note that the constant ρ is cancelled by this division.Likewise, to prove the upper and lower bounds, take the bounds on the area of PN given byLemma 6.15, and divide by the covolume of ΛN from 6.15.We shall now move on to estimating an upper bound for a related function, Uδ(N), also usingsimilar methods.Lemma 6.18. For all δ > 0,Uδ(N) Nα+β+γ−1−δ +Nα +Nβ +Nγ ,where the implicit constant depends only on pa, pb, pc.Proof. By Definition 6.10, for any δ > 0, Uδ(N) counts all triples (a, b, c) ∈ N such that a ≤ Nα−δ,b ≤ Nβ−δ, c ≤ Nγ−δ and pAa a+ pBb = pCc c.LetRa = {(x, y) ∈ R2 : 0 < x ≤ Nα−δ, 0 < y ≤ Nβ},Rb = {(x, y) ∈ R2 : 0 < x ≤ Nα, 0 < y ≤ Nβ−δ},Tc ={(x, y) ∈ R2 : 0 < x, 0 < y, pAapCcx+pBbpCcy ≤ Nγ−δ}.It is clear from the definitions given above that the projection of every (a, b, c) ∈ N3 onto thex, y-plane is contained in the union of Ra, Rb, Tc. Thus it follows by Lemmata 6.14, 6.16 and 6.13thatUδ(N) ≤ hcNγ−1(Volume(Ra) + Volume(Rb) + Volume(Tc))+O(Perimeter(Ra) + Perimeter(Rb) + Perimeter(Tc) + 1).It is clear that Ra, Rb are rectangles and that Volume(Ra) = Nα+β−δ, and Volume(Rb) = Nα+β−δ.Likewise, Tc is a triangle, with vertices clearly at(0, 0),(0,hbhcNβ−δ), and(hahcNα−δ, 0),so Volume(Tc) =hahb2h2cNα+β−2δ. It is geometrically clear that the perimeters have similar magnitudein terms of N as the perimeters of the polygon PN , adjusted at most by constant factors that dependon pa, pb, pc. Thus it follows thatUδ(N) ≤ hahb2hcNα+β+γ−1−2δ + 2hcNα+β+γ−1−δ +O(Nα +Nβ +Nγ).42Which implies thatUδ(N) Nα+β+γ−1−δ +Nα +Nβ +Nγ .Now, we shall follow up with some bounds on some additional functions related to F fromDefinition 6.11.Lemma 6.19. For all d ∈ N thatF (N, d) Nα+β+γ−1d−2,where the implicit coefficient depends only on the choice of primes pa, pb, pc.Proof. For N2 being the set of squarefree natural numbers, letS(N, d) = {(a, b, c) ∈ N32 : pAa a+ pBb b = pCc c, gcd(a, b) = d, a ≤ Nα, b ≤ Nβ , c ≤ Nγ},T (N, d) = {(a, b, c) ∈ N3 : pAa a+ pBb b = pCc c, gcd(a, b) = 1, a ≤ Nα/d, b ≤ Nβ/d, c ≤ Nγ/d},V (N, d) = {(a, b, c) ∈ N3 : pAa a+ pBb b = pC−1c c, gcd(a, b) = 1, a ≤ Nα/d, b ≤ Nβ/d, c ≤ pcNγ/d}.By definition, F (N, d) = #S(N, d).First, we consider the case where d is divisible by a square. Observe that if (a, b, c) ∈ S(N, d)then d divides a, and since d is not squarefree, a is not squarefree, which contradicts the definitionof S(N, d). So it follows S(N, d) = ∅ and thus |F (N, d)| = 0.Now, we move onto the case where d is squarefree, and pc does not divide d. Observe thatif (a, b, c) ∈ S(N, d), since d divides a, b there exists a′, b′ ∈ N such that a/d = a′, b/d = b′ and(a′, b′) = 1. Thus pAa a′d + pBb b′d = pCc c. Since (d, pc) = 1, one has that d divides c and thus thereexists c′ ∈ N such that c/d = c′. Thus by dividing all by d one has thatpAa a′ + pBb b′ = pCc c′; a′ ≤ Nα/d; b′ ≤ Nβ/d; c′ ≤ Nγ/d.Thus (a′, b′, c′) ∈ T (N, d). If one takes two triples (a1, b1, c1), (a2, b2, c2) ∈ S(N, d), observe that(a1/d, b1/d, c1/d) = (a2/d, b2/d, c2/d) if and only if a1 = a2, b1 = b2, c1 = c2. Thus for each(a, b, c) ∈ S(N, d) there exists a unique (a/d, b/d, c/d) ∈ T (N, d) and thusF (N, d) = #S(N, d) ≤ #T (N, d).Observe that for any (a, b, c) ∈ T (N, d), one has thatb ≡ p−Bb pAa a (mod pCc ).43Thus there exists ΨN ∈ N, ΨN ≡ p−Bb pAa a (mod pCc ) with ΨN < pCc . Thus for any (a, b, c) ∈T (N, d), there exists some k ∈ Z≥0 whereb = ΨNa+ pCc k.Thus, it follows thatc =pAapCca+pBbpCcΨNa+ pBb k.Solving for k and using the inequality c ≤ Nγ/d,k ≤ p−Bb Nγ/d−pAa + pBb ΨNpBb pCca≤ hbNβ+γ−1/d.Let K(a) denote the number of possible values k may take for a given a. From the above inequality,it is clear thatK(a) ≤ hbNβ+γ−1/d.By definition, #T (N, d) =∑Nα/da=1 K(a), and thus#T (N, d) =Nα/d∑a=1K(a) ≤Nα/d∑a=1hbNβ+γ−1/d ≤ hbNα+β+γ−1d−2 ≤ pbNα+β+γ−1d−2.Thus it follows that F (N, d) Nα+β+γ−1 when (pc, d) = 1.Finally, we consider the case where d is squarefree, and pc divides d. As in the previous case,obverse that for (a, b, c) ∈ S(N, d), since d divides a, b there exists a′, b′ ∈ N such that a/d =a′, b/d = b′ and (a′, b′) = 1. Thus pAa a′d + pBb b′d = pCc c. As d is squarefree, (d, pCc ) = pc and thusit follows that there exists c′ ∈ N such that pcc = c′d. Thus, dividing by d one has thatpAa a′ + pBb b′ = pC−1c c′; a′ ≤ Nα/d; b′ ≤ Nβ/d; c′ ≤ pcNγ/d.Thus (a′, b′, c′) ∈ V (N, d). It is clear that for every (a, b, c) ∈ S(N, d) there exists a unique triple(a/d, b/d, pcc/d) ∈ V (N, d).ThusF (N, d) = #S(N, d) ≤ #V (N, d).For any (a, b, c) ∈ V (N, d) one has thatb ≡ p−Bb pAa a (mod pC−1c ).44Thus there exists ΨN ∈ N, where ΨN ≡ p−Bb pAa a (mod pC−1c ) and ΨN < pC−1c . Thus for any(a, b, c) ∈ T (N, d), there exists some k ∈ Z≥0 such thatb = ΨNa+ pC−1c k.Thus it follows thatc =pAapC−1ca+pBbpC−1cΨNa+ pBb k.Solving for k with the inequality c ≤ Nγ/d,k ≤ p−Bb Nγ/d−pAa + pBb ΨNpBb pC−1ca≤ hbNβ+γ−1/d.Let K ′(a) denote the number of possible values k may take for a given a. From the above inequality,it is clear thatK ′(a) ≤ hbNβ+γ−1/d.By definition, #V (N, d) =∑Nα/da=1 K′(a), and thus#V (N, d) =Nα/d∑a=1K ′(a) ≤Nα/d∑a=1hbNβ+γ−1/d ≤ pbNα+β+γ−1d−2.Thus it follows that F (N, d) Nα+β+γ−1 when d is squarefree and pc divides d.Lemma 6.20. For all  > 0,F(N) Nα+β+γ−1−,with the implicit constant depending only on pa, pb, pc.Proof. Observe by Lemma 6.19 that for all d ∈ N,F (N, d) Nα+β+γ−1d−2,with the implicit constant depending only on pa, pb, pc.From the definition of F(N) it follows thatF(N) ≤∑d≥N|F (N, d)| ∑d≥NNα+β+γ−1d−2  Nα+β+γ−1∫ ∞Nx−2 dx Nα+β+γ−1−.This completes the proof.456.3 Major Arc StructureThis section deals with the structure of the major arcs and the harmonic arcs contained withinthem. We shall also derive the component of the integral of F (N) that provides what will becomethe asymptotic formula.First, we shall prove a fairly simple statement about the structure of the harmonic major arcs.Though the structure of many of these objects is simple, there are a large number of them, andseveral parameters are required to describe them. So for the sake of clarity, some simple statementssuch as this are given full proofs.Lemma 6.21. Let a, b, c ∈ Z, and qa, qb, qc ∈ N where qa ≤ Qa, qb ≤ Qb, qc ≤ Qb, and (qa, a) = 1,(qb, b) = 1, (qc, c) = 1. If a/(pAa qa) = b/(pBb qb) = c/(pCc qc) then there exists q ∈ N k ∈ Z such thatMα(qa, a) ∩Mβ(qb, b) ∩Mγ(qc, c) = MH(q, k).Furthermore, MH(q, k) ⊆MH provided 0 ≤ c ≤ pCc qc.Proof. If a/(pAa qa) = b/(pBb qb) = c/(pCc qc), it follows that since these three rational numbers areequal, there exist q ∈ N and k ∈ Z such that (q, k) = 1 and q/k = a/(pAa qa) = b/(pBb qb) = c/(pCc qc).Thus it follows thatpAa kq=aqa;pBb kq=bqb;pCc kq=cqc.Since k/q, a/qa, b/qb, c/qc are all in lowest form, it follows from the above equalities thata =pAa k(pAa , q); qa =q(pAa , q); b =pBb k(pBb , q); qb =q(pBb , q); c =pCc k(pCc , q); qc =q(pCc , q).Thus it follows that, by the definition of MH(q, k), thatMα(qa, a)∩Mβ(qb, b) ∩Mγ(qc, c)=Mα(q(q, pAa ),pAa(q, pAa )k)∩Mβ(q(q, pBb ),pBb(q, pBb )k)∩Mγ(q(q, pCc ),pCc(q, pCc )k)=MH(q, k).As (q, k) = 1, to show that MH(q, k) ⊆MH , we must first show that q/(q, pAa ) ≤ Qa, q/(q, pBb ) ≤ Qb,and q/(q, pCc ) ≤ Qc, which holds since by assumption qa ≤ Qa, qb ≤ Qb, qc ≤ Qc. We must secondshow that k ≤ q which holds since 0 ≤ pCc qc = q/k ≤ 1. This completes the proof.Next, we shall demonstrate that an integral over the harmonic arcs can be decomposed intoseveral smaller integrals over the harmonic arc components, in a manner that is analogous to thestandard kind of decomposition of the major arcs found in the trivial point problem.46Lemma 6.22. Suppose 1 ≤ Qa ≤ 12Nα/2, 1 ≤ Qb ≤ 12Nβ/2, 1 ≤ Qc ≤ 12Nγ/2. Let m(q) =min{haQa(q, pAa ), hbQb(q, pBb ), hcQc(q, pCc )}/qN . Let g : R→ C be a function periodic modulo Z.Then for any q ∈ N,q∑k=0(q,k)=1∫MH(q,k)g(x+ k/q) dx = ϕ(q)∫ m(q)−m(q)g(x) dx.Proof. First observe that by definition,MH(q, k) = Mα(q(q, pAa ),pAa(q, pAa )k)∩Mβ(q(q, pBb ),pBb(q, pBb )k)∩Mγ(q(q, pCc ),pCc(q, pCc )k)= {x ∈ [0, 1) : |qpAa x− pAa k| ≤ (q, pAa )QaN−α}∩ {x ∈ [0, 1) : |qpBb x− pBb k| ≤ (q, pBb )QbN−β}∩ {x ∈ [0, 1) : |qpCc x− pCc k| ≤ (q, pCc )QcN−γ}= {x ∈ [0, 1) : |qx− k| ≤ (q, pAa )haQaN−1}∩ {x ∈ [0, 1) : |qx− k| ≤ (q, pBb )hbQbN−1}∩ {x ∈ [0, 1) : |qx− k| ≤ (q, pCc )hcQcN−1}.Thus it follows that by the definition of m(q) given above thatMH(q, k) = {x ∈ [0, 1) : |qx− k| ≤ qm(q)} = {x ∈ [0, 1) : |x− k/q| ≤ m(q)}.Also note that by the bounds on Qa, Qb, Qc, thatqm(q) ≤ min{haNα/2(q, pAa ), hbNβ/2(q, pBb ), hcNγ/2(q, pCc )}/(2N)≤ min{haNα/2pAa , hbNβ/2pBb , hcNγ/2pCc }/(2N)≤ min{N−α/2, N−β/2, N−γ/2}/2 ≤ 1/2. (6.1)1. Suppose q > 1.For any k ∈ Z≥0 where k ≤ q and (q, k) = 1 observe that 1 ≤ k < q since (0, q) > 1 and(q, q) > 1, and thus|q−1| ≤ |k/q|, |q−1| ≤ |1− k/q|.As qm(q) ≤ 1/2 and thus m(q) ≤ q−1/2 < q−1, we have |k/q| > m(q) and |1− k/q| > m(q),so the set {x ∈ [0, 1) : |x− k/q| ≤ m(q)} is the closed interval [k/q−m(q), k/q+m(q)]. Thus47it follows thatq∑k=0(q,k)=1∫MH(q,k)g(x) dx =q∑k=0(q,k)=1∫ k/q+m(q)k/q−m(q)g(x+ k/q) dx=q∑k=0(q,k)=1∫ m(q)−m(q)g(x) dx= ϕ(q)∫ m(q)−m(q)g(x)dx.Thus for q > 1 the lemma holds.2. Suppose q = 1.First note from (6.1) m(1) ≤ 1/2, 0 +m(1) ≤ 1/2 < 1 and 1−m(1) ≥ 1/2 > 0, so it followsthatMH(1, 0) = {x ∈ [0, 1) : |x| ≤ m(1)} = [0,m(1)],andMH(1, 1) = {x ∈ [0, 1) : |x− 1| ≤ m(1)} = [1−m(1), 1).Thus it follows that since g is periodic modulo Z thatq∑k=0(q,k)=1∫MH(q,k)g(x− k/q) dx =∫ m(1)0g(x)dx+∫ 11−m(1)g(x− 1) dx=∫ m(1)0g(x)dx+∫ 0−m(1)g(x) dx=∫ m(1)−m(1)g(x) dx = ϕ(1)∫ m(1)−m(1)g(x) dx = ϕ(q)∫ m(q)−m(q)g(x) dx.Thus for q = 1 the lemma holds.Ultimately, this shows that the harmonic component of the major arcs MH can be decomposedinto a simple integral that does not depend on k. With the next two lemmas, we shall show thatby restricting the values of Qa, Qb, Qc to certain functions of N , we will have that M = MH for Nsufficiently large.Lemma 6.23. Let λa, λb, λc ∈ (0, 1/2], and N sufficiently large.481. Suppose 1 ≤ Qa  Nλaα and 1 ≤ Qb  Nλbβ.Let qa, qb ∈ N, a, b ∈ Z where a/pAa qa 6= b/pBb qb, and qa ≤ Qa, and qb ≤ Qb. If λaα + λbβ <α+ β − 1, then Mα(qa, a) ∩Mβ(qb, b) = ∅.2. Suppose 1 ≤ Qa  Nλaα and 1 ≤ Qc  Nλcγ .Let qa, qc ∈ N, a, c ∈ Z where a/pAa qa 6= c/pCc qc, and qa ≤ Qa, and qc ≤ Qc. If λaα + λcγ <α+ γ − 1, then Mα(qa, a) ∩Mγ(qc, c) = ∅.3. Suppose 1 ≤ Qb  Nλbβ and 1 ≤ Qc  Nλcγ .Let qb, qc ∈ N, b, c ∈ Z where b/pBb qb 6= c/pCc qc, and qb ≤ Qb, and qc ≤ Qc. If λbβ + λcγ <β + γ − 1, then Mβ(qb, b) ∩Mγ(qc, c) = ∅.Proof. It suffices to prove the first case, as the other cases follow by substituting terms. As a/pAa qa 6=b/pBb qb, |a/pAa qa − b/pBb qb| > 0. Since the smallest possible nonzero difference between two integersis 1, it follows that ∣∣∣∣ apAa qa − bpBb qb∣∣∣∣ = 1pAa qapBb qb ∣∣pBb qba− pAa qab∣∣ ≥ 1pAa pBb qaqb .Thus |a/pAa qa − b/pBb qb| ≥ hahbNα+β−2/qaqb. Consequently,Mα(qa, a) ⊆ [a/pAa qa − haQa/qaN, a/pAa qa + haQa/qaN ],Mβ(qb, b) ⊆ [b/pBb qb − hbQb/qbN, b/pBb qb + hbQb/qbN ],so it suffices to show |a/pAa qa − b/pBb qb| > haQa/qaN + hbQb/qbN to prove that Mα(qa, a) ∩Mβ(qb, b) = ∅. Therefore, it suffices to show thatµ =hahbNα+β−2q−1a q−1bhaQa/qaN + hbQb/qbN> 1.Observe that if Qa  Nλaα and Qb  Nλbβ , then since Qa, Qb ≥ 1,µ =hahbNα+β−1haQaqb + hbqaQb≥ hahbNα+β−1(ha + hb)QaQb N (1−λa)α+(1−λb)β−1.Thus by our hypothesis, it follows that for N sufficiently large that µ > 1 and thus Mα(qa, a)∩Mβ(qb, b) = ∅.Now, we use the above lemma to prove that M = MH for N sufficiently large when a specialcondition is satisfied. Recall Definition 6.12 of DH(λa, λb, λc), the harmonic lambda factor, and therelated definition for weakly and strongly harmonic lambdas.49Lemma 6.24. Suppose Qa  Nλaα, Qb  Nλbβ, and Qc  Nλcγ . If DH(λa, λb, λc) is weaklyharmonic, then for N sufficiently large, M = MH .Proof. First, note that since MH ⊆M by definition, it suffices to prove that M ⊆MH .Since maxDH(λa, λb, λc) > 1 (the weakly harmonic condition) then by Lemma 6.23 it followsthat for any qa, qb, qc ∈ N, a, b, c ∈ Z where a/pAa qa 6= b/pBb qb or a/pAa qa 6= c/pCc qc or b/pBb qb 6=c/pCc qc, and qa ≤ Qa, qb ≤ Qb, and qc ≤ Qc, thatMα(qa, a) ∩Mβ(qb, b) ∩Mγ(qc, c) = ∅ ⊂MH .The only other components of M remaining are those where a/pAa qa = b/pbbqb = c/pCc qc ∈ [0, 1],where qa ≤ Qa, qb ≤ Qb, and qc ≤ Qc, a/qa, b/qb, c/qc in lowest form, which are by Lemma 6.21all in MH . Thus it follows that MH = M.The following lemma is a useful tool for translating between different kinds of integral, and willbe used in later sections to compute certain error terms.Lemma 6.25. Let g : R → C be a function periodic modulo Z (that is, g(x) = g(x + n) for anyn ∈ Z and x ∈ R). Then, ∫Mαg(pAa x)dx =∫M?(Qa,Nα)g(x)dx,∫Mβg(pBb x)dx =∫M?(Qb,Nβ)g(x)dx,∫Mγg(−pCc x)dx =∫M?(Qc,Nβg(−x)dx.Proof. It suffices to prove only the first equation. Proofs for the other equations can easily beconstructed by substituting terms.Observe that by Definitions 6.3 and 6.7 of Mα(q, a) and M?(q, a;Qa;Nα) respectively thatx ∈Mα(q, a) if and only if pAa x ∈M?(q, a;Qa;Nα). Thus by the definition of Mα and M?(Qa;Nα),it follows that x ∈Mα if and only if pAa x ∈ (M?(Qa, Nα)+k) for some integer k where 0 ≤ k < pAa .Consequently note that pAa ·Mα, the set produced by pointwise multiplication of the elementsof Mα is precisely the setS =pAa−1⋃n=0(M?(Qa, Nα) + n),50where Mα+n is the set produced by pointwise addition of n. Thus it follows that by the substitutionu = pAa x, ∫Mαg(pAa x) dx = p−Aa∫pAaM?(Qa,Nα)g(u) du= p−AapAa−1∑n=0∫M?(Qa,Nα)+ng(u) du= p−AapAa−1∑n=0∫M?(Qa,Nα)g(u− n) du= p−AapAa−1∑n=0∫M?(Qa,Nα)g(u) du =∫M?(Qa,Nα)g(x) dx.This completes the proof.We now move on to the triple product of the G(q) function, G?3(q). Note that by Definition 6.8that this function is also technically a function of N , and depends on the parameters α, β, γ andpa, pb, pc.We will make use of the following definition in the next lemma:Definition 6.26. Let P ⊆ {pa, pb, pc} be the set such that pa ∈ P if and only if α < 1, and pb ∈ Pif and only if β < 1, and pc ∈ P if and only if γ < 1.Lemma 6.27. The function G?3(q) is a multiplicative function, and for any prime p, and integer` ∈ Z≥0, for N sufficiently large,G?3(p`) =−1/(p2 − 1)3 if p 6∈ P and 1 ≤ ` ≤ 2,1/(p2 − 1)2 if p ∈ P and 1 ≤ ` ≤ 2,0 if ` > 2.Furthermore, when N is sufficiently large, G?3(q)  |G(q)|3, with the sufficiently large conditionand implicit coefficients depending only on pa, pb, pc.Proof. Let Nmax{1−α,1−β,1−γ}/2 ≥ supP . This is our sufficiently large condition. Note the choiceof supP allows for the condition to be trivially true when P = ∅.Let p be any prime. When ` > 2, observe thatG?3(p`) = G(p`/(p`, pAa ))G(p`/(p`, pBb ))G(p`/(p`, pCc )),and as p is relatively prime to at least two of pa, pb, pc and since G(p`) = 0 it follows that G?3(p`) = 0.Thus the rest of the proof need only concern the cases where 0 ≤ ` ≤ 2.51Now let p ∈ P , ` ∈ {1, 2}. Observe that G?3(p`) = G(p`/(p`, pAa ))G(p`/(p`, pBb ))G(p`/(p`, pCc )).Also note that p is equal to one of pa, pb, pc and relatively prime to the other two. Thus if N islarge enough that pa ∈ P implies p2a | pAa , pb ∈ P implies p2b | pBb , and pc ∈ P implies p2c | pCc , itfollows thatG?3(p`) = G(p`)2G(1) = 1/(p2 − 1)2.Now suppose p 6∈ P , ` ∈ {1, 2}. If p = pa then A = 0 and so (p`, pAa ) = 1 regardless of p.Likewise, (p`, pBb ) = 1 and (p`, pCc ) = 1 regardless of p. ThusG?3(p`) = G(p`)3 = −1/(p2 − 1)3.To show G?3(q) |G(q)|3 it suffices to observe that as Nmax{1−α,1−β,1−γ}/2 ≥ supP .G?3(q) = G(q(q, pAa ))G(q(q, pBb ))G(q(q, pCc ))≤ (p2a − 1)(p2b − 1)(p2c − 1)|G(q)|3.We now show that just as the infinite sum∑ϕ(q)G(q)3 = ω, that for N sufficiently large, theinfinite sum∑ϕ(q)G?3(q) stabilizes to a value close to ω.Lemma 6.28. Let P be as in Definition 6.26, and ω the number calculated in Lemma 4.7. For Nsufficiently large,R∑q=1ϕ(q)G?3(q) =∏p∈Pp2 − 1p2 − 2ω +O(R−3/2).Proof. Recall from Lemma 4.7 that the seriesω =∞∑q=1ϕ(q)G(q)3converges and can be rewritten as an Euler product,ω =∏p(1 + φ(p)G(p)3 + ϕ(p2)G(p2)3).For N sufficiently large, Lemma 6.27 implies∞∑q=1φ(q)G?3(q)∞∑q=1φ(q)G(q)3  1,52and thus one can take the Euler product of G?3:∞∑q=1ϕ(q)G?3(1) =∏p(1 + φ(p)G?3(p) + ϕ(p2)G?3(p2))=∏p∈P(1 + ϕ(p)G?3(p) + ϕ(p2)G?3(p2))∏p 6∈P(1 + ϕ(p)G?3(p) + 1 + ϕ(p2)G?3(p2))=∏p∈P(1 +p− 1(p2 − 1)2 +p(p− 1)(p2 − 1)2)∏p 6∈P(1 + ϕ(p)G(p)3 + ϕ(p2)G(p2)3)=∏p∈P(1 +1p2 − 1)(1 + ϕ(p)G(p)3 + ϕ(p2)G(p2)3)−1ω= ω∏p∈P(1 +1p2 − 1)(1− 1(p2 − 1)2)−1= ω∏p∈Pp2 − 1p2 − 2 .Observing that by Lemmata 6.27 and 4.6, for N sufficiently large,∑q>Qϕ(q)G?3(q)∑q≥Q|ϕ(q)G(q)3|  Q−3/2,so we haveω∏p∈P(p2 − 1p2 − 2)=∞∑q=1ϕ(q)G?3(q) =Q∑q=1ϕ(q)G?3(q) +O(Q−3/2).The proof follows immediately on rearrangement.Finally, we use the previous lemmas in this section to calculate what will eventually become theasymptotic formula of F (N).Lemma 6.29. If Q0 = min{Qa, Qb, Qc}, then∫MHf?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx = ζ(2)−3∑q≤Q0ϕ(q)G?3(q)U(N) +O(Nα+β+γ−1Q−3/20 ).Proof. Let Q˜ = {q ∈ N : q ≤ min{(q, pAa )Qa, (q, pBb )Qb, (q, pCc )Qc}}, so thatMH =⋃q∈Q˜q⋃k=0(q,k)=1MH(q, k).53Next, we break the integral of MH into two components,∫MHf?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx=∑q≤Q0q∑k=0(q,k)=1∫MH(q,k)f?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx+∑q>Q0q∈Q˜q∑k=0(q,k)=1∫MH(q,k)f?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx=∑q≤Q0q∑k=0(q,k)=1∫MH(q,k)f?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx+ EH ,where EH is the second double sum in the centre expression.Definem(q) = min{haQa(q, pAa ), hbQb(q, pBb ), hcQc(q, pCc )}/(qN).Observe that by Lemma 6.27 if G?3(q) 6= 0, m(q) ≤ max{p2a, p2b , p2c}h0Q0/qN , so m(q) = O(Q0/qN).Also note that regardless of G?3(q),m(q)−1 = qN max{1haQa(q, pAa ),1hbQb(q, pBb ),1hcQc(q, pCc )}≤ qN/Q0,so it follows that m(q)−1  qN/Q0.For fixed q ∈ Q˜, we apply Lemma 6.22 withg(x− k/q) = Iα(pAa x− pAa k/q)Iβ(pBb x− pBb k/q)Iγ(pCc x− pCc k/q)to show that:q∑k=0(q,k)=1∫MH(q,k)f?α(pAa x)f?β(pBb x)f?γ (−pCc x)dx= ζ(2)−3ϕ(q)G(q(q, pAa ))G(q(q, pBb ))G(q(q, pCc ))∫ m(q)−m(q)Iα(pAa x)Iβ(pBb x)Iγ(−pCc x)dx= ζ(2)−3ϕ(q)G?3(q)(∫ 12− 12Iα(pAa x)Iβ(pBb x)Iγ(−pCc x)dx+O(∫ 12m(q)‖p−Aa p−Bb p−Cc x−3‖dx))= ζ(2)−3ϕ(q)G?3(q)(∫ 10Iα(pAa x)Iβ(pBb x)Iγ(−pCc x)dx+O(Nα+β+γ−3∫ 12m(q)‖x‖−3dx))= ζ(2)−3ϕ(q)G?3(q)(U(N) +O(Nα+β+γ−3m(q)−2))= ζ(2)−3ϕ(q)G?3(q)(U(N) +O(q2Nα+β+γ−1Q−20 )).54Thus by Lemmata 6.27 and 4.6 one has that for all  > 0:∑q≤Q0ϕ(q)G?3(q)(q2Nα+β+γ−1Q−20 )∑q≤Q0q3|G(q)|3Nα+β+γ−1Q−20 Nα+β+γ−1Q−20∑q≤Q0q3|G(q)|3 Nα+β+γ−1+Q−3/20 .Thus it follows that∫MHf?α(pAa x)f?β(pBb x)f?γ (−pCc x)dx=∑q≤Q0(ζ(2)−3φ(q)G?3(q)U(N))+O(Nα+β+γ−1Q−3/20 ) + EH ,Thus to prove the lemma, it now suffices to show that EH  Nα+β+γ−1Q−3/20 . First, observethat for any q ∈ Q˜, if G?3(q) 6= 0 it follows by Lemma 6.27 that (q, pAa ) ≤ p2a, (q, pBb ) ≤ p2b , (q, pCc ) ≤p2c , so q ≤ p20Q0 and q  Q0. Consequently it follows by Lemma 6.27 thatEH ≤ ζ(2)−3∑Q0<q≤p20Q0ϕ(q)G?3(q)∫ m(q)−m(q)Iα(x)Iβ(x)Iγ(x)dx∑Q0<q≤p20Q0qG?3(q)Nα+β+γm(q)∑Q0<q≤p20Q0|G(q)|3Nα+β+γ−1Q0  Nα+β+γ−1Q0∑Q0<q≤p20Q0|G(q)|3.Now observe that by Lemma 4.6, one has that for all  > 0∑Q0<q≤p20Q0|G(q)|3  Q−5/2+0 .Combining the two relations gives the result that EH  Nα+β+γ−1Q−3/20 .The formula given above can be simplified further, leading to the following, final lemma of thissection.Lemma 6.30. Let Q0 = min{Qa, Qb, Qc} and P be as Definition 6.26. Then∫MHf?α(pAa x)f?β(pBb x)f?γ (−pCc x)dx = ωζ(2)−3∏p∈Pp2 − 1p2 − 2U(N) +O(Nα+β+γ−1Q−3/20 ).Proof. This follows from Lemmata 6.28 and 6.29, and observing that the hypotheses for both aresatisfied once N is sufficiently large, and removing the smaller error term.55The above lemma contains the core of the integral, the component that will be the main term inthe future. The magnitude of the main term is Nα+β+γ−1, which can be seen by combining it withLemma 6.17. Observe that the error on this term is always strictly smaller than the main term aslong as there exists some δ > 0 such that Q0  Nδ. Practically speaking, this will always hold inthis work.6.4 Bounds of Functions on Component ArcsWe now move on to a discussion of some technical lemmas which will be used to determine the sizesof various error terms.First, we estimate the delta functions over their component major arcs.Lemma 6.31. For all  > 0, ∫Mα|∆α(pAa x)|3dx Nα+Q2a,∫Mβ|∆β(pBb x)|3dx Nβ+Q2b ,∫Mγ|∆γ(−pCc x)|3dx Nγ+Q2c .Proof. The proof for the first equation is easily generalized to the other two equations, so it sufficesto prove only the first one. By Lemma 6.25,∫Mα|∆α(pAa x)|3dx =∫M?(Qa,Nα)|∆α(x)|3dx.First, note that as fα, f?α are exponential sums with Nα +O(1) terms,|∆α(x)| = |fα(x)− f?α(x)| ≤ |fα(x)|+ |f?α(x)|  Nα,so sup |∆α(x)|  Nα. By application of Lemma 3.2 from [2], after taking the sup-norm one hasthat for any  > 0,∫M?(Qa,Nα)|∆α(x)|3dx ≤ supx∈M?(Qa,Nα)|∆α(x)|∫M?(Qa,Nα)|∆α(x)|2 dx Nα+Q2a.This completes the proof.The above bound for the triple ∆ component may not be ideal, and, as will be seen later, thisbound will be one of the most significant contributors to the size of the error terms.Next, we estimate the f? functions over their component arcs.56Lemma 6.32. For all  > 0, ∫Mα|f?α(pAa x)|3dx N2α,∫Mβ|f?β(pBb x)|3dx N2β ,∫Mγ|f?γ (−pCc x)|3dx N2γ .Proof. The proof of the three equations is symmetric, so it suffices to prove only the first one. ByLemma 6.25, ∫Mα|f?α(pAa x)|3dx =∫M?(Qa,Nα)|f?α(x)|3dx.Note that f?α(x) is an exponential sum with Nα +O(1) terms so sup |f?α(x)|  Nα. By applicationof Lemma 4.1 from [2], after taking the sup-norm one has that for any  > 0,∫M?(Qa,Nα)|f?α(x)|3dx ≤ supx∈M?(Qa,Nα)|f?α(x)|∫M?(Qa;Nα)|f?α(x)|2dx N2α(1 +Q−1/2+a ).This completes the proof.The f? bounds are the best possible bounds, as we know the exponential sums get large onportions of the arc.Next, we provide a function for converting from our minor arc components to minor arcs asdefined in [2].Lemma 6.33. Let g : R→ C be a function periodic modulo Z. Then∫mαg(pAa x) dx =∫m?(Qa;Nα)g(x) dx,∫mβg(pBb x) dx =∫m?(Qb;Nβ)g(x) dx,∫mγg(−pCc x) dx =∫m?(Qc;Nγ)g(−x) dx.Proof. Since the proof of each equation is symmetric, it suffices to prove only the first. Observethat by Lemma 6.25 that ∫Mαg(pAa x) dx =∫M?(Qa,Nα)g(x) dx.Also note that by substitution of u = pAa x that,∫ 10g(pAa x) dx = p−Aa∫ pAa0g(u) du =∫ 10g(x) dx.57Now, observe that∫mαg(pAa x) dx =∫ 10g(pAa x)dx−∫Mαg(pAa x) dx=∫ 10g(x)dx−∫M?(Qa;Nα)g(x) dx =∫m?(Qa;Nα)g(x) dx.This completes the proof.We now provide a bound on certain products of functions related to G(q). This is much like theearlier proof for G?3(q), but only an upper bound is needed as these only appear in certain errorterms.Lemma 6.34. Uniformly as a function of N ,G(q(q, pAa ))G(q(q, pBb )) |G(q)|2,G(q(q, pAa ))G(q(q, pCc )) |G(q)|2,G(q(q, pBb ))G(q(q, pCc )) |G(q)|2.Proof. It suffices to prove only the first equation as the proofs for the other two are symmetric. Forany q, let q′ ∈ N, ka, kb ∈ Z≥0 whereq = pkaa pkbb q′; gcd(pa, q′) = gcd(pb, q′) = 1.If ka > 2 then G(q/(q, pBb )) = 0. If kb > 2 then G(q/(q, pAa )) = 0.Thus if G(q/(q, pAa ))G(q/(q, pBb )) 6= 0, then ka ≤ 2, kb ≤ 2. Thus p2ap2bq′ ≥ q, so|G(q/(q, pAa ))G(q/(q, pBb ))| ≤ |G(q′)|2 ≤ |G(q)|2(|G(p2a)G(p2b)|)−1.Since |G(p2a)G(p2b)| is a constant, it follows that G(q/(q, pAa ))G(q/(q, pBb )) |G(q)|2.Just as a bound for products of two values of G will be needed for a component of the errorterm, an estimate for an integral of a pair of I functions over the unit interval will be needed aswell.Lemma 6.35. For all  > 0,∫ 10|Iα(pAa x)Iβ(pBb x)|2 dx N2α+2β−1 +Nα+β ,∫ 10|Iα(pAa x)Iγ(pCc x)|2 dx N2α+2γ−1 +Nα+β ,∫ 10|Iβ(pBb x)Iγ(pCc x)|2 dx N2β+2γ−1 +Nα+β .58Proof. It suffices to prove only the first equation as the arguments for the second and third aresymmetric. Observe that∫ 10|Iα(pAa x)Iβ(pBb x)|2 dx is clearly a counting function which counts theelements of the setS = {(x, y, z, w) ∈ N : pAa x+ pBb y = pAa z + pBb w, x, z ≤ Nα, y, w ≤ Nβ}.Observe that for any (x, y, z, w) ∈ S, one has that pAa x ≡ pAa z (mod pBb ), and since pa, pb arerelatively prime, it follows that x ≡ z (mod pBb ). Thus z = x + pBb k for some k ∈ Z. Similarly,y ≡ w (mod pAa ), so there exists ` ∈ Z such that w = y + pAa `.Rewriting in terms of k, ` instead of z, w, the equation becomespAa x+ pBb y = pAa x+ pAa pBb k + pBb y + pAa pBb `,and thus after simplifying it follows that ` = −k. So one can represent (x, y, z, w) ∈ S uniquelywith just (x, y) and k. Observe that as 1 ≤ z ≤ Nα, 1 ≤ x+ pBb k ≤ Nα, so1− xpBb≤ k ≤ hbNα+β−1 − xpBbIf α+ β− 1 > 0, there are hbNα+β−1 +O(1) possible values for k to take. For x, y, there are atmost Nα+β possible values to take, so therefore #S  N2α+2β−1. If α + β ≤ 1, there is only onevalue of k, k = 0 that is valid. Hence #S = Nα+β . The desired result follows immediately.We now move on an observation about the structure of the semi-major arcs that will allow forsubstantial reduction of the error term.Lemma 6.36. Suppose there exist λa, λb, λc ∈ [0, 1/2] such thatQa  Nλaα; Qb  Nλbβ ; Qc  Nλcγ .If DH(λa, λb, λc) is strongly harmonic, ∂MH = ∂M.Proof. First observe by Definition 6.6 that ∂MH is trivially a subset of ∂M, since ∂MH is a unionof subsets of ∂M, so to prove equality it suffices to show that ∂M ⊆ ∂MH .If minDH(λa, λb, λc) > 1 then every inequality in Lemma 6.23 is satisfied. Thus for all qa, qb, qc ∈N, and a, b, c ∈ Z, where (qa, a) = (qb, b) = (qc, c) = 1 and a ≤ pAa qa, and b ≤ pBb qb, and c ≤ pCc qc,one has thatMα(qa, a) ∩Mβ(qb, b) 6= ∅ iff apAa qa=bpBb qbMα(qa, a) ∩Mγ(qc, c) 6= ∅ iff apBa qa=cpCc qcMβ(qb, b) ∩Mγ(qc, c) 6= ∅ iff bpBb qb=cpCc qC.59Since by definition, ∂M is a subset of the union of the sets of one of the three the forms above,and those sets are non-empty if and only if they lie in the harmonic semi-majors, it follows that∂M ⊆ ∂MH , thus completing the proof.This suffices as a set of tools to resolve most of the components of the error term that remain.We shall now move on to actually calculating the various components of the error term.6.5 Triple ∆ Error TermsIn this section, we compute the error terms over components of the major, semi-major, and semi-minor arcs that have the form∫ |∆α(pAa x)∆β(pBb x)∆γ(pCc x)| dx.All the proofs in this section are straightforward applications of Ho¨lder’s inequality to decomposethe integral into components resolvable by lemmata in the previous sections, followed by a finallemma which summarizes the overall result.First, we compute them over the major arcs:Lemma 6.37. For all  > 0,∫M|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx N (α+β+γ)/3+Q2/3a Q2/3b Q2/3c .Proof. Recall that M ⊆Mα. Thus by Ho¨lder’s inequality∫M|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx≤(∫Mα|∆α(pAa x)|3dx∫Mβ|∆β(pBb x)|3dx∫Mγ|∆γ(−pCc x)|3dx)1/3.To complete the proof, simply apply Lemma 6.31 to the above equation.Second, we compute them over the semi-major arcs:Lemma 6.38. For all  > 0 one has that∫∂M|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx N (α+β+γ)/3+Q2/3a Q2/3b (Nγ/3Q−1/2c +Q1/3c )+N (α+β+γ)/3+Q2/3a Q2/3c (Nβ/3Q−1/2b +Q1/3b )+N (α+β+γ)/3+Q2/3b Q2/3c (Nα/3Q−1/2a +Q1/3a ).60Proof. First observe that by the definition of the semimajor arcs,∫∂M|∆α(pAa x)∆β(pBb x)∆γ(pccx)|dx=∫Mα∩Mβ∩mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx+∫Mα∩mβ∩Mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx+∫mα∩Mβ∩Mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx.Thus by Ho¨lder’s inequality, and Lemmata 6.25, 6.33, 5.1, and 6.31 that:∫Mα∩Mβ∩mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx≤(∫Mα|∆α(pAa x)|3dx∫Mβ|∆β(pBb x)|3dx∫mγ|∆γ(pCc x)|3dx)1/3(Nα+β+Q2aQ2b∫m?(Qc;Nγ)|f(x;Nγ)|3dx)1/3 N (α+β)/3+Q2/3a Q2/3b (N2γQ−3/2c +Qc)1/3 N (α+β+γ)/3+Q2/3a Q2/3b (Nγ/3Q−1/2c +Q1/3c ).To complete the proof, apply the same lemmas to the other two components of the integral.Second, we compute them over the semi-minor arcs:Lemma 6.39. For all  > 0,∫∂m|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (α+β+γ)/3+Q2/3a (Nβ/3Q−1/2b +Q1/3b )(Nγ/3Q−1/2c +Q1/3c )+N (α+β+γ)/3+Q2/3b (Nα/3Q−1/2a +Q1/3a )(Nγ/3Q−1/2c +Q1/3c )+N (α+β+γ)/3+Q2/3c (Nα/3Q−1/2a +Q1/3a )(Nβ/3Q−1/2b +Q1/3b ).61Proof. First observe that by the definition of the semi-minor arcs,∫∂m|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx=∫Mα∩mβ∩mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx+∫Mβ∩mα∩mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx+∫Mγ∩mα∩mβ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx.Now observe that by Ho¨lder’s inequality, followed by Lemmata 6.25, 6.33, 5.1, and 6.31 that:∫Mα∩mβ∩mγ|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx≤(∫Mα|∆α(pAa x)|3 dx∫mβ|∆β(pBb x)|3 dx∫mγ|∆γ(−pCc x)|3 dx)1/3(Nα+Q2a∫m?(Qb;Nβ)|f(x;Nβ)|3 dx∫m?(Qb;Nγ)|f(x;Nγ)|3 dx)1/3(Nα+Q2a(N2β+Q−3/2b +Nβ+Qb)(N2γ+Q−3/2c +Nγ+Qc))1/3 N (α+β+γ)/3+Q2/3a (Nβ/3Q−1/2b +Q1/3b )(Nγ/3Q−1/2c +Q1/3c ).To complete the proof, apply the above lemmas to the other two components of the arc.Finally, we compute them over the minor arcs:Lemma 6.40. For all  > 0,∫m|∆α(pAa x;Nα)∆β(pBb x;Nβ)∆γ(−pCc x;Nγ)| dx N (α+β+γ)/3+(Nα/3Q−1/2a +Q1/3a )(Nβ/3Q−1/2b +Q1/3b )(Nγ/3Q−1/2c +Q1/3c ).Proof. First note that as the minor arcs is exactly the subset of [0, 1) that is minor for all threecomponents,∫m|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx =∫m|f(pAa x;Nα)f(pBb x;Nβ)f(pCc x;Nγ) dx.62Observe that by Ho¨lder’s Inequality, Lemma 6.33, and Lemma 5.1, that for all  > 0,∫m|f(pAa x;Nα)fβ(pBb x;Nβ)fγ(−pCc x;Nγ)| dx≤(∫m|f(pAa x;Nα)|3 dx∫m|f(pBb x;Nβ)|3 dx∫m|f(pCc x;Nγ)|3)1/3dx≤(∫m?(Qa,Nα)|f(x;Nα)|3 dx∫m?(Qb,Nβ)|f(x;Nβ)|3 dx∫m?(Qc,Nγ)|f(x;Nγ)|3 dx)1/3 (N2α+Q−3/2a +Nα+Qa)1/3(N2β+Q−3/2b +Nβ+Qb)1/3(N2γ+Q−3/2c +Nγ+Qc)1/3 N (α+β+γ)/3+(Nα/3Q−1/2a +Q1/3a )(Nβ/3Q−1/3b +Q1/3b )(Nγ/3Q−1/2c +Q1/3b ).By combining all these results, we get following bound for the integral over the entire interval:Lemma 6.41. For all  > 0,∫ 10|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (α+β+γ)/3+(Nα/3Q−1/2a +Q2/3a )(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ).If we choose λa, λb, λc so thatQa = Nλaα; Qb = Nλbβ ; Qc = Nλcγ ;then this error term is optimal when λa = λb = λc = 2/7. This optimal bound exceeds the mainterm Nα+β+γ−1 when α+ β + γ ≤ 21/10.Proof. The above equation follows immediately by combining the results of Lemmata 6.37, 6.38,6.39, 6.40. Consequently, if Qa = Nλaα, Qb = Nλbβ , Qc = Nλcγ , then it follows by that λa = λb =λc = 2/7 is the point where the error term is minimal. Above this threshold, the error increaseswith λa, λb, λc.Thus, at best, ∫ 10|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N11(α+β+γ)/21.To observe the limit on the error, combine the above equation with the observation that1121(α+ β + γ) < α+ β + γ − 1 if and only if α+ β + γ > 2110.63While the error given above can be made smaller than the main term for α+ β + γ > 2.1, it isclear that the minor arc component alone has a far superior ideal bound. If one sets Qa = N2α/5,Qb = N2β/5 and Qc = N2γ/5, by Lemma 6.40 the bound on the minor arcs for the triple-deltacomponent isN7(α+β+γ)/15+,which is strictly smaller than the main term when α+ β + γ > 1.875.The primary cause of the larger size in the earlier terms comes from the relatively large boundon∫ |∆(x)|3 dx. If a superior technique were found to optimize the bound on this component alone,it is plausible that the minor arcs could become the limiting factor.6.6 Single f ?, Double ∆ Error TermsWe now move on to error terms of the form∫ |f?∆∆| dx.As in the previous section, we shall find bounds for the errors over the major, semi-major, andsemi-minor arcs. Over the minor arcs, errors of this form are zero, as the f? functions are truncatedto be zero over the minor arcs.Unlike the previous section, some of these proofs involve slightly more complex machinery. Thetechnique used in these proofs is similar to the one used to prove Lemma 4.2 in [2]. As these willuse common notation, we shall provide it here, rather than repeat it over multiple lemmas.64For any R ∈ R where 1 ≤ R ≤ Qa, letσb =logQblogQa, σc =logQclogQa,Ra = R, Rb = Rσb , Rc = Rσc ,Nα(q, a;R) = {x ∈ [0, 1) : |qpAa x− a| ≤ Ra/Nα},Nβ(q, b;R) = {x ∈ [0, 1) : |qpBb x− b| ≤ Rb/Nβ},Nγ(q, c;R) = {x ∈ [0, 1) : |qpCc x− c| ≤ Rc/Nγ},Nα(R) =⋃q≤Raq⋃a=1(q,a)=1Nα(q, a;R),Nβ(R) =⋃q≤Rbq⋃b=1(q,b)=1Nβ(q, b;R),Nγ(R) =⋃q≤Rcq⋃c=1(q,c)=1Nγ(q, c;R),Pα(R) = Nα(2R) \Nα(R).Observe that Nα,Nβ ,Nγ are just rescalings of the major arc components by some power of R,so any lemmas that apply to major arcs can apply to their rescalings.First, the major arcs:Lemma 6.42. For all  > 0,∫M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (2α+β+γ)/3+(1 +Q−1/2a Q2/3b Q2/3c )∫M|∆α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx N (α+2β+γ)/3+(1 +Q2/3a Q−1/2b Q2/3c )∫M|∆α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx N (α+β+2γ)/3+(1 +Q2/3a Q2/3b Q−1/2c ).Proof. It suffices to prove the first equation, as the others follow by similar logic. Observe thatNα(R),Nβ(R),Nγ(R) are the major arc components Mα,Mβ ,Mγ with Qa, Qb, Qc rescaled toRa, Rb, Rc respectively. Also observe that the sets{Pα(R) : 1 ≤ R ≤ Qa/2}; {Pβ(R) : 1 ≤ R ≤ Qa/2}; {Pγ(R) : 1 ≤ R ≤ Qa/2};each cover Mα, Mβ , and Mγ respectively. It is also clear that it takes O(logQa) distinct R to covereach. Combining this dyadic dissection with Ho¨lder’s inequality,∫M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx log(2Qa) max1≤R≤Qa/2(U1(R)U2(R)U3(R))1/3,65whereU1(R) =∫Pα(R)|f?α(pAa x)|3 dx;U2(R) =∫Nβ(2R)|∆β(pBb x)|3 dx;U3(R) =∫Nγ(2R)|∆γ(pCc x)|3 dx.By Lemma 6.31, for all  > 0,∫Nβ(R)|∆β(pBb x)|3 dx Nβ+R2b ;∫Nγ(R)|∆(pCc )|3 dx Nγ+R2c .By Lemma 6.25, ∫Pα(R)|f?α(pAa x)|3 dx =∫M?(2R;Nα)\M?(R;Nα)|f?α(x)|3 dx.We expand the above expression in a manner similar to that found in Lemma 4.2 of [2]. Observethat Pα(R) is the set of major arcs Nα(q, k; 2R) on the interval for q ∈ (R, 2R], but is a union ofannular regions of the form Nα(q, k; 2R) \Nα(q, k;R), q ∈ [1, R], soU1(R) =∫Pα(R)|f?α(pAa x)|3 dx=∑R<q≤2Rφ(q)|G(q)|3∫ 2R/qNα−2R/qNα|Iα(x)|3 dx+∑q≤Rφ(q)|G(q)|3∫ 2R/qNαR/qNα|Iα(x)|3 dx,and follow with the application of Lemma 4.6 to give us that for all  > 0,U1(R)∑R<q≤2Rφ(q)|G(q)|3∫ 1/2−1/2|Iα(x)|3 dx+∑1≤q≤Rφ(q)|G(q)|3∫ 1/2R/qNα‖x‖3 dx N2α ∑R<q≤2Rq|G(q)|3+N2αR−2 ∑1≤q≤Rq3|G(q)|3 N2αR−3/2+ +N2αR−3/2+.By combining these results, it follows thatmax1≤R≤Qa/2(U1(R)U2(R)U3(R))1/3  N (2α+β+γ)/3+(1 +Q−1/2a Q2/3b Q2/3c ),and thus ∫M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)|dx N (2α+β+γ)/3+(1 +Q−1/2a Q2/3b Q2/3c ).66Now, the semimajor arcs.Lemma 6.43. For all  > 0,∫∂M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (2α+β+γ)/3+((1 +Q−1/2a Q2/3b )(Nγ/3Q−1/2c +Q1/3c )+ (1 +Q−1/2a Q2/3c )(Nβ/3Q−1/2b +Q1/3b ))∫∂M|∆α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx N (α+2β+γ)/3+((1 +Q2/3a Q−1/2b )(Nγ/3Q−1/2c +Q1/3c )+ (1 +Q−1/2b Q2/3c )(Nα/3Q−1/2a +Q1/3a ))∫∂M|∆α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx N (α+β+2γ)/3+((1 +Q2/3a Q−1/2c )(Nβ/3Q−1/2b +Q1/3b )+ (1 +Q2/3b Q−1/2c )(Nα/3Q−1/2a +Q1/3a )).Proof. It suffices to prove the first equation as the others can be proven by similar logic. Observethat Nα(R),Nβ(R),Nγ(R) are the major arc components Mα,Mβ ,Mγ with Qa, Qb, Qc rescaled toRa, Rb, Rc respectively. Also observe that the sets{Pα(R) : 1 ≤ R ≤ Qa/2}; {Pβ(R) : 1 ≤ R ≤ Qa/2}; {Pγ(R) : 1 ≤ R ≤ Qa/2};each cover Mα, Mβ , and Mγ respectively. It is also clear that it takes O(logQa) distinct R to covereach.Now, note that if |f?α(pAa x)| 6= 0 then x ∈ Mα, and that implies for x ∈ ∂M, that x ∈ Mα ∩Mβ ∩mγ or x ∈Mα ∩mβ ∩Mγ . Thus by Ho¨lder’s inequality,∫∂M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x) dx≤(∫Mα∩Mβ|f?α(pAa x)∆β(pBb x)|3/2 dx)2/3(∫mγ|∆γ(pCc x)|3dx)1/3+(∫Mα∩Mγ|f?α(pAa x)∆γ(pCc x)|3/2 dx)2/3(∫mβ|∆γ(pBb x)|3dx)1/3.Observe that ∆ = f on the minor arcs. Combining this with the dyadic dissection above and67applying Cauchy’s inequality,∫∂M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx log(2Qa) max1≤R≤Qa/2(U1(R)U2(R))1/3(∫mγ|f(pCc x;Nγ)|3 dx)1/3+ log(2Qa) max1≤R≤Qa/2(U1((R)U3(R))1/3(∫mβ|f(pBb x;Nβ)|3 dx)1/3,whereU1(R) =∫Pα(R)|f?α(pAa x)|3 dx;U2(R) =∫Nβ(2R)|∆β(pBb x)|3 dx;U3(R) =∫Nγ(2R)|∆γ(pCc x)|3 dx.By Lemma 6.31, for all  > 0,∫Nβ(R)|∆β(pBb x)|3 dx Nβ+R2b ;∫Nγ(R)|∆(pCc )|3 dx Nγ+R2c .By Lemmata 6.33 and 5.1,∫mβ|f(pBb x;Nβ)|3 dx NβQ−3/2b +Qb;∫mγ|f(pCc x;Nγ)|3 dx NγQ−3/2c +Qc.By Lemma 6.25, ∫Pα(R)|f?α(pAa x)|3dx =∫M?(2R;Nα)\M?(R;Nα)|f?α(x)|3 dx.By the same technique described in the previous lemma, for all  > 0,U1(R)∑R<q≤2Rφ(q)|G(q)|3∫ 1/2−1/2|Iα(x)|3 dx+∑1≤q≤Rφ(q)|G(q)|3∫ 1/2R/qNα‖x‖3 dx N2α ∑R<q≤2Rq|G(q)|3+N2αR−2 ∑1≤q≤Rq3|G(q)|3 N2αR−3/2+ +N2αR−3/2+.By combining these results, we havemax1≤R≤Qa/2(U1(R)U2(R))1/3  N (2α+β)/3+(1 +Q−1/2a Q2/3b ),max1≤R≤Qa/2(U1(R)U3(R))1/3  N (2α+γ)/3+(1 +Q−1/2a Q2/3c ).and thus ∫M|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (2α+β)/3+(1 +Q−1/2a Q2/3b )(Nγ/3Q−1/2c +Q1/3c )+N (2α+γ)/3+(1 +Q−1/2a Q2/3c )(Nβ/3Q−1/2b +Q1/3b ).68We now move on to the error term of this form over the semi-minor arcs. Unlike the earlierproofs, there is nothing to gain by breaking the arcs into dyadic intervals, so this is a straightforwardapplication of Ho¨lder’s inequality.Lemma 6.44. For all  > 0,∫∂m|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q1/3b )(Nγ/3Q−1/2c +Q1/3c ),∫∂m|∆α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx N (α+2β+γ)/3+(NαQ−1/2a +Q1/3a )(Nγ/3Q−1/2c +Q1/3c ),∫∂m|∆α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx N (α+β+2γ)/3+(Nα/3Q−1/2a +Q1/3a )(Nβ/3Q−1/2b +Q1/3b ).Proof. It suffices to prove the first equation as the others can be proven by symmetric logic.First note that f?(pAa x) 6= 0 implies that x ∈Mα. By the definition of the semi-minor arcs, thisimplies x ∈ mβ ∩mγ . Thus observe that by Ho¨lder’s inequality,∫∂M|f?(pAa x)∆β(pBb x)∆γ(−pCc x)| dx≤(∫Mα|f?(pAa x)|3 dx∫mβ|∆β(pBb x)|3 dx∫mγ|∆γ(pCc x)|3 dx)1/3.By Lemmata 6.33 and 5.1, for all  > 0∫mβ|∆β(pBb x)|3 dx N2βQ−3/2b +NβQb;∫mγ|∆γ(pCc x)|3 dx N2γQ−3/2c +NγQc.By Lemma 6.32, ∫Mα|f?α(pAa x)|3 dx N2α.Thus it follows that∫∂M|f?(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N2α/3+(N2β/3Q−1/2b +Nβ/3Q1/3b )(N2γ/3Q−1/2c +Nγ/3Q1/3c ),completing the proof.This leads to the final result of the section:69Lemma 6.45. For all  > 0,∫ 10|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ),∫ 10|∆α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx N (α+2β+γ)/3+(Nα/2Q−1/2a +Q2/3a )(Nγ/2Q−1/2c +Q2/3c ),∫ 10|∆α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx N (α+β+2γ)/3+(Nα/2Q−1/2a +Q2/3a )(Nβ/2Q−1/2b +Q2/3b ).If there exists some λa, λb, λc such thatQa = Nλaα; Qb = Nλbβ ; Qc = Nλcγ ,then the error terms for all three equations are optimal when λa = λb = λc = 2/7. This optimalbound is strictly smaller than the main term when the following inequalities are satisfied:0.7α+ β + γ > 2.1; α+ 0.7β + γ > 2.1; α+ β + 0.7γ > 2.1.Proof. The above equations follow immediately by combining Lemmata 6.42, 6.43, and 6.44 withthe observation that on the minor arcs, the integrals are zero since all f? functions are zero there.When Qa  Nλaα, Qb  Nλbβ λa = λb = λc = 2/7, the error term is of the form∫ 10|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx N2α/3+11(β+γ)/21+,∫ 10|∆α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx N2β/3+11(α+γ)/21+,∫ 10|∆α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx N2γ/3+11(α+β)/21+.Observe that as2α/3 + 19(β + γ)/21 < α+ β + γ − 1 iff α/3 + 10(β + γ)/21 > 1,2β/3 + 19(α+ γ)/21 < α+ β + γ − 1 iff β/3 + 10(α+ γ)/21 > 1,2γ/3 + 19(α+ β)/21 < α+ β + γ − 1 iff γ/3 + 10(α+ β)/21 > 1.the region stated in the proof follows immediately.Observe that the bound is primarily determined by the contribution from the semi-minor arcs.6.7 Harmonic Arc Error TermsBefore moving to the actual double f?, single ∆ error terms, we shall first investigate the behaviourof these integrals on the harmonic majors, and on an extension of the concept of the harmonicmajors to the semi-major arcs.70First, we prove a result about the double f?, single ∆ error terms on the major arcs. This usesa technique similar to the one used in the previous section, but with some changes, so it is fullyexplained within the proof of the lemma.Lemma 6.46. Let Q0 = min{Qa, Qb, Qc}. For all  > 0,∫MH|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β−1/2+,∫MH|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx Nα+γ−1/2+,∫MH|∆α(pAa x)f?β(pBb x)f?γ (−pCc x)| dx Nβ+γ−1/2+,provided α+ β > 1, α+ γ > 1, and β + γ > 1 for each equation respectively.Proof. Let X ⊆ [0, 1) be the set such that f?α(pAa x)f?β(pBb x)∆γ(−pCc x) 6= 0.For any R such that 1 ≤ R ≤ Q0:σa =logQalogQ0, σb =logQblogQ0, σc =logQclogQ0,Ra = Rσa , Rb = Rσb , Rc = Rσc ,m(q;R) = min{haRa(q, pAa )qN,hbRb(q, pBb )qN,hcRc(q, pCc )qN},Q˜(R) = {q ∈ N : q ≤ Ra(q, pAa ), q ≤ Rb(q, pBb ), q ≤ Rc(q, pCc )},Nα(q, a;R) = {x ∈ [0, 1) : |qpAa x− a| ≤ Ra/Nα},Nβ(q, b;R) = {x ∈ [0, 1) : |qpBb x− b| ≤ Rb/Nβ},Nγ(q, c;R) = {x ∈ [0, 1) : |qpCc x− c| ≤ Rc/Nγ},N(q, k;R) = Nα(q(q, pAa ),pAa(q, pAa )k;R) ∩Nβ( q(q, pBb ),pBb(q, pBb )k;R) ∩Nγ( q(q, pCc ),pCc(q, pCc )k;R),N(R) =⋃q∈Q˜(R)q⋃k=1(q,k)=1N(q, k;R),P(R) = N(2R) \N(R).The above definitions are directly analogous to the definition of the harmonic majors, withQa, Qb, Qc rescaled. Observe that by definition, MH can be covered by the set of dyadic intervals{P(R) : R ∈ [1, Q1/2]},71and since the number of dyadic intervals needed to cover MH is proportional to logQ0 it thusfollows that∫MH|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx log(2Q0) max1≤R≤Q0(U1(R)U2(R))1/2,whereU1(R) =∫P(R)|f?α(pAa x)f?β(pBb x)|2 dx; U2(R) =∫N(2R)∩X|∆γ(pCc x)|2 dx.Observe that if N(q, k; 2R) ⊆ N(2R) ∩ X, this implies that q ≤ min{p2aRa, p2bRb, p2cRc}, which isless than q ≤ max{p2a, p2b , p2c}R. For K = max{p2a, p2b , p2c},N(2R) ∩X ⊆KR⋃q=1q⋃c=1(q,c)=1{x ∈ [0, 1) : |qpCc x− c| ≤ KR/Nγ}.From Lemma 3.2 of [2] we have a bound on the L2 norm of ∆ type functions over the appropriatemajor arc. As this set is of the form of Mγ but with a different maximum, it follows by Lemma6.25 that for all  > 0, thatU2(R) =∫M?(2R;Nγ)|∆γ(pCc x)|2 dx R2+.To estimate U1(R), first observe that for q ∈ Q˜(2R) \ Q˜(R) that if G(q/(q, pAa ))G(q/(q, pBb )) 6= 0then Qa < q ≤ 2p2aQa, Qb < q ≤ 2p2bQb and Qc < q ≤ 2p2cQc.By Lemma 6.34, G(q/(q, pAa )G(q/(q, pBb )) G(q)2.Let p1 = max{pa, pb, pc}, Thus for U1(R), observe that by the definition of P(~R), Lemma 6.35,72and the fact that α+ β > 1 that∫P(R)|f?α(pAa x)f?β(pBb x)|2 dx=∑q∈Q˜(2R)\Q˜(R)q∑k=1(q,k)=1)∫N(q,k;2R)|f?α(pAa x)f?β(pBb x)|2 dx+∑q∈Q˜(R)q∑k=1(q,k)=1∫N(q,k;2R)\N(q,k,R)|f?α(pAa x)f?β(pBb )|2 dx≤∑q∈tildeQ(2R)\Q˜(R)φ(q)|G(q/(q, pAa ))G(q/(q, pBb )|2∫ 10|Iα(pAa x)Iβ(pBb )|2 dx+∑q∈Q˜(R)φ(q)|G(q/(q, pAa ))G(q/(q, pBb ))|2∫ 1/2m(q;R)|Iα(pAa x)Iβ(pBb )|2 dx∑R<q≤2p20Rq|G(q)|4(N2α+2β−1 +Nα+β) +∑q≤Rq|G(q)|4∫ 1/2m(q;R)|Iα(pAa x)Iβ(pBb x)|2 dx∑R<q≤2p20Rq|G(q)|4N2α +∑q≤Rq|G(q)|4p−2Aa p−2Bb∫ 1/2m(q;R)‖x‖−4 dx∑R<q≤2p20Rq|G(q)|4N2α +∑q≤Rq|G(q)|4N2α+2β−4m(q;R)−3 N2α+2β−1∑R<q≤2p20Rq|G(q)|4 +N2α+2β−1R−3∑q≤R0q4|G(q)|4.By Lemma 4.6 it follows that∑R0<q≤2p1R0q|G(q)|4  R−5/2+;∑q≤Rq4|G(q)|4  R1/2+,and thusU1(R) N2α+2β−1R−5/2.Hence, it follows that ∫MH|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx log(2Q0) max1≤R≤Q0/2(N2α+2β−1+R−5/2R2)1/2 log(2Q0) max1≤R≤Q0/2(N2α+2β−1+R−1/2)1/2 Nα+β−1/2+.This completes the proof.73Note that the condition that α+β > 1, α+γ > 1, and β+γ > 1 are there simply to ensure thatthe bound from Lemma 6.35 is of the Nα+β−1/2 form and not the full Nα+β−1/2 +Nα/2+β/2 form.This restriction is not particularly cumbersome, and the main term cannot actually be resolvedcompletely unless all three of the above are satisfied, since when one is not satisfied the error termof U(N) from Lemma 6.17 becomes too large. Removing the condition produces an equally usablelemma, but with a more cumbersome to write error term. This observation will apply to the lastlemma of this section as well.We now move on to the semi-major arcs. Recall from Definition 6.6 the harmonic semimajorarcs, and their low and high components. We shall now bound the error terms on these arcs forintegrals with two f? functions and one ∆ function.We first compute the low harmonic semi-major arcs. These are annular regions near the majorarcs.Lemma 6.47. For all  > 0, it follows that∫∂MH0|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β+γ/2−1/2+Q−7/4c Q1/40 +Nα+β−1/2+Q−1/2c Q1/40 ,∫∂MH0|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx Nα+β/2+γ−1/2+Q−7/4b Q1/40 +Nα+γ−1/2+Q−1/2b Q1/40 ,∫∂MH0|∆α(pAa x)f?β(pBb x)f?γ (−pCc x)| dx Nα/2+β+γ−1/2+Q−7/4a Q1/40 +Nβ+γ−1/2+Q−1/2a Q1/40 .Proof. Observe that if pCc x ∈Mγ , this implies one ofpAa x 6∈Mα, pBb x 6∈Mβ ,and thus |f?α(pAa x)f?β(pBb x)| = 0. Thus it follows that∫∂MH0|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx =∫∂MH0∩mγ|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx.Thus it follows that∫∂MH0∩mγ|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx=∑q∈∂Q˜q∑k=1(q,k)=1∫∂MH(q,k)∩mγ|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx.74Thus it follows by Cauchy’s inequality that∫∂MH0∩mγ|f?α(pAa x)f?β(pBb x)∆γ(pCc )| dx≤(∫∂MH0|f?α(pAa x)f?β(pBb x)|2 dx∫mγ|∆γ(pCc x)|2 dx)1/2≤(∫∂MH0|f?α(pAa x)f?β(pBb x)|2 dx∫mγ|f(pCc x;Nγ)|2 dx)1/2.By Lemma 6.33, followed by Theorem 2 of [2] one has that:∫mγ|f(−pCc x;Nγ)|2 dx =∫m?(Qc,Nγ)|fγ(x)|2 dx NγQ−1/2c +Q2c .Letmc(q) =hcQc(q, pBb )qN.Observe that ∂MH0 ∩mγ is a union of annular regions around each major arc, and when one fixesq, each annular region is bounded within [K − 1/2,K −mc(q)] ∪ [mc(q) +K, 1/2 +K] where K isthe centre of the annulus. Thus by Lemma 6.34,∫∂MH0∩mγ|f?α(pAa x)f?β(pBb x)|2 dx=∑q∈Q˜q∑k=1(q,k)=1∫∂MH(q,k)∩mγ|f?α(pAa x)f?β(pBb x)|2 dx∑q∈Q˜φ(q)|G(q/(q, pAa ))G(q/(q, pBb ))|2∫ 1/2mc(q)|Iα(pAa x)Iβ(pBb x)|2 dx∑q∈Q˜q|G(q/(q, pAa ))G(q/(q, pBb ))|2∫ 1/2mc(q)p−2Aa p−2Bb ‖x‖−4 dx∑q∈Q˜q|G(q/(q, pAa ))G(q/(q, pBb ))|2N2α+2β−4mc(q)−3∑q≤p20Q0q|G(q)|4N2α+2β−4q3N3Q−3c N2α+2β−1Q−3c∑q∈∂Q˜q4|G(q)|4.Thus by Lemma 4.6, ∫∂MH0|f?α(pAa x)f?β(pBb x)|2 dx N2α+2β−1+Q−3c Q1/20 ,75and therefore,∫∂MH0|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β+γ/2−1/2+Q−7/4c Q1/40 +Nα+β−1/2+Q−1/2c Q1/40 .This completes the proof for this case. As the others are symmetric, this suffices.Second, we compute the high harmonic semi-major arcs. These are the remains of the harmonicarcs after q becomes large.Lemma 6.48. Let Q0 = min{Qa, Qb, Qc}. For all  > 0,∫∂MH1|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β−1/2+(Nγ/2Q−3/2c +Q−1/4c ),∫∂MH1|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx Nα+γ−1/2+(Nβ/2Q−3/2b +Q−1/4b ),∫∂MH1|∆α(pAa x)f?β(pBb x)f?γ (−pCCx)| dx Nβ+γ−1/2+(Nα/2Q−3/2a +Q−1/4a ),when α+ β > 1, α+ γ > 1, and β + γ > 1 respectively.Proof. We shall prove this for the first equation only, as the other two equations can be proven bysymmetric logic. First, by Cauchy’s inequality, and the fact that f?α and f?β are non-zero only (interms of the semimajor arcs) on the intersection of Mα ∩Mβ ∩mγ ,∫∂MH1|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx≤(∫∂MH1|f?α(pAa x)f?β(pBb x)|2 dx∫mγ|∆γ(−pCc x)|2 dx)1/2.First, the minor arc integral is resolved by observing that by Lemma 6.33 followed by Theorem 2of [2], ∫mγ|∆γ(pCc x)|2 dx =∫m?(Qc;Nγ)|∆γ(x)|2 dx NγQ−1/2c +Q2c .Now, we resolve the remaining integral. Let∂Q˜H = {q ∈ N : exactly two of the inequalitiesq ≤ Qa(q, pAa ), q ≤ Qb(q, pBb ), q ≤ Qc(q, pCc ) hold}.76Observe that by Lemmata 6.34, 6.35, and 4.6, it follows that∫∂MH1|f?α(pAa x)f?β(pBb x)|2 dx≤∑q∈Q˜Hq∑k=0(q,k)=1∫∂M(q,k)|f?α(pAa x)f?β(pBb x)|2 dx≤∑q∈Q˜Hϕ(q)|G(q/(q, pAa ))G(q/(q, pBb ))|2∫ 1/2−1/2|Iα(pAa x)Iβ(pBb x)|2 dx∑q∈Q˜Hq|G(q)|4(N2α+2β−1 +Nα+β) N2α+2β−1∑q≥Qcq|G(q)|4 N2α+2β−1Q−5/2c .The proof is completed by combining the two components with the earlier inequality.This completes this section, and prepares all the tools needed to compute the double f?, single∆ integral error terms.6.8 Double f ?, Single ∆ Error TermsWe shall now combine the above lemmas about behaviour on the harmonics with our understandingof the relation between the harmonic major and semimajor arcs and the entire major and semimajorarcs to give use the error terms derived from the integrals with two f? and one ∆ component.First, the major arcs:Lemma 6.49. Fix λa, λb, λc ∈ [0, 1/2] soQa  Nλaα; Qb  Nλbβ ; Qc  Nλcγ .If DH(λa, λb, λc) is weakly harmonic, then for all  > 0,∫M|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β−1/2+,∫M|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx Nα+γ−1/2+,∫M|∆α(pAa x)f?β(pBb x)f?γ (−pCc x)| dx Nβ+γ−1/2+.Proof. This follows immediately from the application of Lemma 6.24 to Lemma 6.46.77Second, the semimajor arcs:Lemma 6.50. Fix λa, λb, λc ∈ [0, 1/2] soQa  Nλaα; Qb  Nλbβ ; Qc  Nλcγ .If DH(λa, λb, λc) is strongly harmonic, then for all  > 0,∫∂M|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β−1/2+(Nγ/2Q−3/2c +Q−1/4c ),∫∂M|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx Nα+γ−1/2+(Nβ/2Q−3/2b +Q−1/4b ),∫∂M|∆α(pAa x)f?β(pBb x)f?γ (−pCc x)| dx Nβ+γ−1/2+(Nα/2Q−3/2a +Q−1/4a ).Proof. We shall give an explicit proof for the first equation only. By Lemmata 6.47 and 6.48,∫∂MH|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β−1/2+((Nγ/2Q−7/4c +Q−1/2c )Q1/40 + (Nγ/2Q−3/2c +Q−1/4c )) Nα+β−1/2+(Nγ/2Q−3/2c +Q−1/4c ).By the strongly harmonic condition and Lemma 6.36, it follows by that ∂MH = ∂M, completingthe proof.As the semiminor and minor arcs have at least two minor components, any error term with twof? components must be zero as f? is by definition zero on the minor arcs, so we can now move onto the error over the whole integral:Lemma 6.51. LetQ0 = min{Qa, Qb, Qc}.Let λa, λb, λc ∈ [0, 1/2] such thatQa  Nλaα; Qb  Nλbβ ; Qc  Nλcγ .If DH(λa, λb, λc) is strongly harmonic, it follows that for all  > 0,∫ 10|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx Nα+β−1/2+(1 +Nγ/2Q−3/2c ),∫ 10|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx Nα+γ−1/2+(1 +Nβ/2Q−3/2b ),∫ 10|∆α(pAa x)f?β(pBb x)f?γ (−pCc x)| dx Nβ+γ−1/2+(1 +Nα/2Q−3/2a ).78Proof. This follows immediately from Lemmata 6.49 and 6.50 and the observation that on thesemiminor and minor arcs, the integrals with two f? components must be zero as each x in thosearcs is in at most one major arc component.With this, we can conclude our work on the 1f?2∆ type integrals, and move on to the compu-tation of the whole integral.6.9 Computation of the IntegralIn this section, and the next, we shall unite all the errors from the various cross products of f? and∆ into a single function of Qa, Qb, Qc.Definition 6.52. Let us define the error termE(Qa, Qb, Qc;N) =∫ 10f(pAa x;Nα)f(pBb x;Nβ)f(−pCc x;Nγ) dx−∫ 10f?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx,recalling that the definition of the f? functions depend on Qa, Qb, Qc.From our earlier results, we construct a bound on E.Lemma 6.53. Suppose there exist λa, λb, λc ∈ [0, 1/2] such that Qa  Nλaα, Qb  Nλbβ, Qc Nλcγ . If DH(λa, λb, λc) is strongly harmonic, then for all  > 0,E(Qa, Qb, Qc;N) N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c )+N (α+2β+γ)/3+(Nα/3Q−1/2a +Q2/3b )(Nγ/3Q−1/2c +Q2/3c )+N (α+β+2γ)/3+(Nα/3Q−1/2a +Q2/3a )(Nβ/3Q−1/2b +Q−2/3b )+Nα+β−1/2+(1 +Nγ/2Q−3/2c )+Nα+γ−1/2+(1 +Nβ/2Q−3/2b )+Nβ+γ−1/2+(1 +Nα/2Q−3/2a ).79Proof. First, we observe that by definition,E(Qa, Qb, Qc;N)=∫ 10f(pAa x;Nα)f(pBb x;Nβ)f(−pCc x;Nγ)dx−∫ 10f?α(pAa x)f?β(pBb x)f?γ (−pCc x)dx=∫ 10(∆α(pAa x)− f?α(pAa x))(∆β(pAa x)− f?β(pBb x))(∆γ(−pCc x)− f?γ (−pCc x)) dx−∫ 10f?α(pAa x)f?β(pBb x)f?γ (−pCc x) dx≤∫ 10|∆α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx+∫ 10|f?α(pAa x)∆β(pBb x)∆γ(−pCc x)| dx+∫ 10|∆α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx+∫ 10|∆α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx+∫ 10|f?α(pAa x)f?β(pBb x)∆γ(−pCc x)| dx+∫ 10|f?α(pAa x)∆β(pBb x)f?γ (−pCc x)| dx+∫ 10|∆α(pAa x)f?β(pBb x)f?γ (−pCc x)| dx.Thus by application of Lemmata 6.41, 6.45, and 6.51, one has that provided the conditions forLemma 6.51 are satisfied, that for all  > 0,E(Qa, Qb, Qc;N) N (α+β+γ)/3+(Nα/3Q−1/2a +Q2/3a )(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c )+N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c )+N (α+2β+γ)/3+(Nα/3Q−1/2a +Q2/3b )(Nγ/3Q−1/2c +Q2/3c )+N (α+β+2γ)/3+(Nα/3Q−1/2a +Q2/3a )(Nβ/3Q−1/2b +Q−2/3b )+Nα+β−1/2+(1 +Nγ/2Q−3/2c )+Nα+γ−1/2+(1 +Nβ/2Q−3/2b )+Nβ+γ−1/2+(1 +Nα/2Q−3/2a ).This gives us the general bound for this lemma, after noting that the first term is strictly smallerthan the next three for any λ ∈ [0, 1/2], and removing the strictly smaller terms.From this, we determine the geometric description of regions where it is possible to ensure thatE(Qa, Qb, Qc)Lemma 6.54. Let λa, λb, λc ∈ (0, 1/2]. Let Qa = Nλa , Qb = Nλb , Qc = Nλc . Letηa = min{1 + 3λa/2, 2− 2λa}; ηb = min{1 + 3λb/2, 2− 2λb}; ηc = min{1 + 3λc/2, 2− 2λc},νa = min{1 + 3λa, 2}; νb = min{1 + 3λb, 2}; νc = min{1 + 3λc, 2}.80For each (α, β, γ) ∈ R(λa, λb, λc) there exists some δ > 0 such thatE(Qa, Qb, Qc;N) Nα+β+γ−1−δ,where R(λa, λb, λc) is the intersection of the following volumes:V1 = {(α, β, γ) ∈ [0, 1]3 : (1− λa)α+ (1− λb)β > 1},V2 = {(α, β, γ) ∈ [0, 1]3 : (1− λa)α+ (1− λc)γ > 1},V3 = {(α, β, γ) ∈ [0, 1]3 : (1− λb)β + (1− λc)γ > 1},V4 = {(α, β, γ) ∈ [0, 1]3 : α+ ηbβ + ηcγ > 3},V5 = {(α, β, γ) ∈ [0, 1]3 : ηaα+ β + ηcγ > 3},V6 = {(α, β, γ) ∈ [0, 1]3 : ηaα+ ηbβ + γ > 3},V7 = {(α, β, γ) ∈ [0, 1]3 : νcγ > 1},V8 = {(α, β, γ) ∈ [0, 1]3 : νbβ > 1},V9 = {(α, β, γ) ∈ [0, 1]3 : νaα > 1}.Proof. Observe that in order to satisfy the conditions for Lemma 6.53, we must have that(1− λa)α+ (1− λa)β > 1, α(1− λa) + γ(1− λc) > 1, β(1− λb) + γ(1− λc) > 1,which can be represented geometrically as a set of volumes V1, V2, V3 with corresponding planarboundaries pi1, pi2, pi3:V1 = {(α, β, γ) : (1− λa)α+ (1− λb)β > 1},V2 = {(α, β, γ) : (1− λa)α+ (1− λc)γ > 1},V3 = {(α, β, γ) : (1− λb)β + (1− λc)γ > 1},pi1 = {(α, β, γ) : (1− λa)α+ (1− λb)β = 1},pi2 = {(α, β, γ) : (1− λa)α+ (1− λc)γ = 1},pi3 = {(α, β, γ) : (1− λb)β + (1− λc)γ = 1}.Let V = V1 ∩ V2 ∩ V3. By Lemma 6.53, for all (α, β, γ) ∈ V and all  > 0,E(Nλa , Nλb , Nλc ;N) N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c )+N (α+2β+γ)/3+(Nα/3Q−1/2a +Q2/3a )(Nγ/3Q−1/2c +Q2/3c )+N (α+β+2γ)/3+(Nα/3Q−1/2a +Q2/3a )(Nβ/3Q−1/2b +Q2/3b )+Nα+β−1/2+(1 +Nγ/2Q−3/2c )+Nα+γ−1/2+(1 +Nβ/2Q−3/2b )+Nβ+γ−1/2+(1 +Nα/2Q−3/2a ).81It is clear from the above that there are two distinct forms of terms, with Qa, Qb, Qc and α, β, γpermuted for a total of six terms.First, we evaluate the geometric space on whichN (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ) is strictly less than Nα+β+γ−1, by splitting the evaluation into four cases:1. If λb ≤ 2/7 and λc ≤ 2/7,N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ) N (2α+β+γ)/3+(Nβ/3−λbβ/2+γ/3−λcγ/2) N2(α+β+γ)/3−λbβ/2−λcγ/2+,which implies that the condition that the error be smaller than the main term is2(α+ β + γ)/3− λbβ/2− λcγ/2 < α+ β + γ − 11− λbβ/2− λcγ/2 < (α+ β + γ)/33 < α+ (1 + 3λb/2)β + (1 + 3λc/2)γ,and thus the volume V4 where the error term is smaller than the main term isV4 = {(α, β, γ) : α+ (1 + 3λb/2)β + (1 + 3λc/2)γ > 3},with corresponding boundary planepi4 = {(α, β, γ) : α+ (1 + 3λb/3)β + (1 + 3λc/2)γ = 3}.2. If λb ≤ 2/7 and λc > 2/7,N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ) N (2α+β+γ)/3+(Nβ/3−λbβ/2+2λcγ/3) N (2α+2β+γ)/3−λbβ/2+2λcγ/3+,which implies that the condition that the error be smaller than the main term is(2α+ 2β + γ)/3− λbβ/2 + 2λcγ/3 < α+ β + γ − 11− λbβ/2 + 2λcγ/3 < (α+ β + 2γ)/33 < α+ (1 + 3λb/2)β + (2− 2λc)γ,and thus the volume V4 where the error term is smaller than the main term isV4 = {(α, β, γ) : α+ (1 + 3λb/2)β + (2− 2λc)γ > 3},with corresponding boundary planepi4 = {(α, β, γ) : α+ (1 + 3λb/3)β + (2− 2λc)γ = 3}.823. If λb > 2/7 and λc ≤ 2/7,N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ) N (2α+β+γ)/3+(N2λbβ/3+γ/3−λcγ/2) N (2α+β+2γ)/3+2λbβ/3−λcγ/2+,which implies that the condition that the error be smaller than the main term is(2α+ β + 2γ)/3 + 2λbβ/2− λcγ/2 < α+ β + γ − 11 + 2λbβ/3− λcγ/2 < (α+ 2β + γ)/33 < α+ (2− 2λb)β + (1 + 3λc/2)γ,and thus the volume V4 where the error term is smaller than the main term isV4 = {(α, β, γ) : α+ (2− 2λb)β + (1 + 3λc/2)γ > 3},with corresponding boundary planepi4 = {(α, β, γ) : α+ (2− 2λb)β + (1 + 3λc/2)γ = 3}.4. If λb > 2/7 and λc > 2/7:N (2α+β+γ)/3+(Nβ/3Q−1/2b +Q2/3b )(Nγ/3Q−1/2c +Q2/3c ) N (2α+β+γ)/3+(N2λbβ/3+2λcγ/3) N (2α+β+γ)/3+2λbβ/3+2λcγ/3+,which implies that the condition that the error be smaller than the main term is(2α+ β + γ)/3 + 2λbβ/3 + 2λcγ/3 < α+ β + γ − 11 + 2λbβ/3 + 2λcγ/3 < (α+ 2β + 2γ)/33 < α+ (2− 2λb)β + (2− 2λc)γ,and thus the volume V4 where the error term is smaller than the main term isV4 = {(α, β, γ) : α+ (2− 2λb)β + (2− 2λc)γ > 3},with corresponding boundary planepi4 = {(α, β, γ) : α+ (2− 2λb)β + (2− 2λc)γ = 3}.83By our choice of ηa, ηb, ηc, it is clear that this can be simplified so thatV4 = {(α, β, γ) : α+ ηbβ + ηcγ > 3},with boundarypi4 = {(α, β, γ) : α+ ηbβ + ηcγ = 3}.As there are two error terms that are merely permutations of the previous, it is clear that theyare strictly smaller than Nα+β+γ−1 on the corresponding volumesV5 = {(α, β, γ) : ηaα+ β + ηcγ > 3},V6 = {(α, β, γ) : ηaα+ ηbβ + γ > 3}.with corresponding boundariespi5 = {(α, β, γ) : ηaα+ β + ηcγ = 3},pi6 = {(α, β, γ) : ηaα+ ηbβ + γ = 3}.We shall now consider the geometric space where Nα+β−1/2+(1+Nγ/2Q−3/2c ) is strictly smallerthan Nα+β+γ−1. We split this problem into two cases:1. If λc ≥ 1/3,Nα+β−1/2+(1 +Nγ/2Q−3/2c ) = Nα+β−1/2+,so it follows the condition that the error be smaller than the main term isα+ β − 1/2 < α+ β + γ − 1,1/2 < γ,and thus the volume V7 where the error is smaller than the main term isV7 = {(α, β, γ) : 2γ > 1},with boundarypi7 = {(α, β, γ) : 2γ = 1}.2. If λc ≤ 1/3,Nα+β−1/2+(1 +Nγ/2Q−3/2c = Nα+β+γ/2−3λcγ/2−1/2+,so it follows the condition that the error be smaller than the main term isα+ β + γ/2− 3λcγ/2− 1/2 < α+ β + γ − 1,1/2 < γ/2 + 3λcγ/2,1 < (1 + 3λc)γ,84and thus the volume V7 where the error is smaller than the main term isV7 = {(α, β, γ) : (1 + 3λc)γ > 1},with boundarypi7 = {(α, β, γ) : (1 + 3λc)γ = 1}.By our choice of νa, νb, νc, it is clear that this simplifies toV7 = {(α, β, γ) : νcγ > 1},with boundarypi7 = {(α, β, γ) : νcγ = 1}.The two remaining error terms are permutations of this one, so it follows the corresponding volumeson which those error terms are smaller than Nα+β+γ−1 areV8 = {(α, β, γ) : νbβ > 1},V9 = {(α, β, γ) : νaα > 1},with boundariespi8 = {(α, β, γ) : νbβ = 1},pi9 = {(α, β, γ) : νaα = 1}.Combining all these together gives the geometric object described by the lemma.Let us define the following volume:Definition 6.55. Let R be the following union of volumes:R =⋃λa∈(0,1/2]⋃λb∈(0,1/2]⋃λc∈(0,1/2]R(λa, λb, λc),where R(λa, λb, λc) is as defined in Lemma 6.54.Observe that by Lemma 6.54, it is clear that R is precisely the set of parameters (α, β, γ) suchthat there exist λa, λb, λc whereE(Nλa , Nλb , Nλc ;N) Nα+β+γ−1−δfor some δ > 0.We shall now give a geometric description of R:85Lemma 6.56. The volumeR =⋃2/7≤λ≤1/3(R(λ, 2/7, 2/7) ∪R(2/7, λ, 2/7) ∪R(2/7, 2/7, λ)).which contains the non-empty open volume bounded by0.7α+ β + γ > 2.1; α+ 0.7β + γ > 2.1; α+ β + 0.7γ > 2.1;13α > 7; 13β > 7; 13γ > 7;α = 1; β = 1; γ = 1,and is contained in the volume0.7α+ β + γ > 2.1; α+ 0.7β + γ > 2.1; α+ β + 0.7γ > 2.1;α > 1/2; β > 1/2; γ > 1/2;α = 1; β = 1; γ = 1.Proof. First, we describe the geometric shape of R(λa, λb, λc) for some fixed λa, λb, λc ∈ (0, 1/2].Let V1, . . . , V9 be the volumes described by Lemma 6.54. Let ηa, ηb, ηc and νa, νb, νc be as definedin Lemma 6.54. Let pi1, . . . , pi9 be the planes bounding the corresponding volume.1. The rectangular prism X = V7 ∩ V8 ∩ V9.By the definition of the three volumes, it is clear they form a rectangular prism:X = (1νa, 1]× ( 1νb, 1]× ( 1νc, 1].Furthermore, it is clear that X = (1/2, 1]3 if and only if λa, λb, λc ≥ 1/3, with X ⊆ (1/2, 1]3otherwise.2. The triangular pyramid Y = V4 ∩ V5 ∩ V6.Observe that the boundary pi4 ∩ pi5 forms the line`1 =(1ηa − 1 ,1ηb − 1 ,ηaηb − 1ηc(ηa − 1)(ηb − 1))t+ (0, 0, 3),and likewise pi4 ∩ pi6 forms the line`2 =(1ηa − 1 ,ηaηc − 1ηb(ηa − 1)(ηc − 1) ,1ηc − 1)t+ (0, 3, 0),and finally pi5 ∩ pi6 forms the line`3 =(ηbηc − 1ηa(ηb − 1)(ηc − 1) ,1ηb − 1 ,1ηc − 1)t+ (3, 0, 0).86These three lines intersect at the pointP = 3(ηa − 1)(ηb − 1)(ηc − 1) ·(1ηa(ηb − 1)(ηc − 1) + (ηa − 1)(ηbηc − 1) ,1ηb(ηa − 1)(ηc − 1) + (ηb − 1)(ηaηc − 1) ,1ηc(ηa − 1)(ηb − 1) + (ηc − 1)(ηaηb − 1)).Furthermore, it is clear that since ηa, ηb, ηc ≤ 10/7 (with equality if and only if λ = 2/7),V4 ⊆W4 = {(α, β, γ) : α+ 10β/7 + 10γ/7 > 3},V5 ⊆W5 = {(α, β, γ) : 10α/7 + β + 10γ/7 > 3},V6 ⊆W6 = {(α, β, γ) : 10α/7 + 10β/7 + γ > 3},and that by the relation between η and λ as one moves away from maximum values one hasthat if|λa − 2/7| ≥ |λ′a − 2/7|, |λb − 2/7| ≥ |λ′b − 2/7|, |λc − 2/7| ≥ |λ′c − 2/7|,then for j = 4, 5, 6,Vj(λ′a, λ′b, λ′c) ⊆ Vj(λa, λb, λc).Let Z = V1∩V2∩V3. When 2/7 ≤ λa, λb, λc ≤ 1/3, one has that X ∩Y ⊆ Z since in this range,V1 ⊆ V6, V2 ⊆ V5 and V3 ⊆ V4.We also know from manner in which X and Y shrink as the λ parameters move from 2/7 and1/3 that it is necessarily the case thatR =⋃2/7≤λa≤1/3⋃2/7≤λb≤1/3⋃2/7≤λc≤1/3R(λa, λb, λc),and that furthermore,R =⋃2/7≤λ≤1/3(R(λ, 2/7, 2/7) ∪R(2/7, λ, 2/7) ∪R(2/7, 2/7, λ)).The non-empty open volume given by the proof is the volume of R(2/7, 2/7, 2/7), which isclearly the majority of R. The upper bound on the volume is given by the union of the individualmaximums for X and Y in the range 2/7 to 1/3.Now, with a bound for the error term, when α, β, γ lies in R, it is possible to construct anasymptotic formula for the function F (N).87Theorem 6.57. Let P be as in Definition 6.26. If (α, β, γ) ∈ R, then there exists some δ > 0 suchthatF (N) = ωζ(2)−3∏p∈Pp2 − 1p2 − 2U(N) +O(Nα+β+γ−1−δ),and specifically,F (N) ≤ ωζ(2)−3∏p∈Pp2 − 1p2 − 2Nα+β+γ−1 +O(Nα+β+γ−1−δ),F (N) ≥ ω2pcζ(2)−3∏p∈Pp2 − 1p2 − 2Nα+β+γ−1 +O(Nα+β+γ−1−δ).Proof. By Lemma 6.24, for any (α, β, γ) ∈ R, there exist λa, λb, λc ∈ (0, 1/2] such thatminDH(λa, λb, λc) > 1 and M = MH .Let Qa = Nλaα, Qb = Nλbβ , Qc = Nλcγ . By definition,F (N) =∫ 10f?α(pAa x)f?β(pBb x)f?γ (−pCc x)dx+ E(Qa, Qb, Qc;N).Since for any x ∈ [0, 1), f?α(pAa x) 6= 0 if and only if x ∈ Mα, f?β(pBb x) 6= 0 if and only if x ∈ Mβ ,and f?γ (−pCc x) 6= 0 if and only if x ∈Mγ , it follows thatF (N) =∫Mf?α(pAa x)f?β(pBb x)f?γ (−pCc x)dx+ E(Qa, Qb, Qc;N).As M = MH and (α, β, γ) ∈ R, by Lemmata 6.30 and 6.56, it follows that there exists some δ > 0such thatF (N) = ωζ(2)−3∏p∈Pp2 − 1p2 − 2U(N) +O(Nα+β+γ−1−δ.To complete this proof, simply factor in the bounds on U(N) as determined in Lemma 6.17.Note that the error terms of U(N) can be accounted for by making δ small enough thatα+ β + γ − 1− δ > max{α, β, γ},which is possible since for all (α, β, γ) ∈ R, α+ β + γ − 1 > max{α, β, γ}.At the end of all this, we have an asymptotic formula which is valid for all parameters (α, β, γ) ∈R. However, we are not yet done. Our initial problem dealt with the limit points in the set Q. Asdiscussed at the beginning of §6, this is not enough to prove immediately that R ⊆ Q′ as F (N) doesnot distinguish between relatively prime triples and places no restriction on the relative sizes of thefirst two numbers, so it does not count only abc-triples. In the next section, we prove Theorem 2.2.886.10 Proof of Theorem 2.2Here, we show that the closure of R (a volume whose bounds given in Lemma 6.56) lies in Q′. AsQ′ is a closed set, it suffices to show that R lies in Q′.For any δ > 0 letSδ(N) = {(a, b, c) ∈ N3 : pAa a+ pBb b = pCc c, gcd(a, b) ≥ Nδ,Nα−δ < a ≤ Nα, Nβ−δ < b ≤ Nβ , Nγ−δ < c < Nγa, b, c are squarefree}.Let F ?δ (N) = #(Sδ(N)).Recall Uδ(N) from Definition 6.10 and F(N) from Definition 6.11. For any δ > 0 one has thatF ?δ (N) ≥ F (N)− Uδ(N)− Fδ(N).If (α, β, γ) ∈ R2/7, by Theorem 6.57, one has that there exists some  > 0 such thatF (N) ≥ ω2pcζ(2)−3∏p∈Pp2 − 1p2 − 2Nα+β+γ−1 +O(Nα+β+γ−1−).where P is the set described in Definition 6.26, and N is sufficiently large.It follows from Lemmata 6.18 and 6.20 that for all δ > 0F ?δ (N) ≥ω2pcζ(2)−3∏p∈Pp2 − 1p2 − 2Nα+β+γ−1 +O(Nα+β+γ−1−min{,δ}).for N sufficiently large. Thus for every δ > 0 there exists some Mδ ∈ R such that for all N ≥ Mδ,Sδ(N) is non-empty.In the rest of the proof we take A,B,C to depend in the usual way on N , but with N =max1≤m≤n{M1/m} + n. That is to say, A,B,C depend on n. From Mδ above, there exists asequence of triples T whereT = {(pAa an, pBb bn, pCc cn)}∞n=1, (an, bn, cn) ∈ S1/n( max1≤m≤n{M1/n}+ n).First, observe that for (pAa an, pBb bb, pCc cn) ∈ T there is some dn = gcd(an, bn) where dn <(max1≤m≤n{M1/m} + n)1/n. Let a′n = an/dn, b′n = bn/dn, and c′n = gcd(pc, dn)cn/dn. As dndivides an, dn is squarefree, so gcd(pc, dn) = gcd(pCc , dn), and thus it follows thatpAa a′n + pBb b′n ∈ {pC−1c c′n, pCc c′n}.89Thus for N = max1≤m≤n{M1/m}+ nNα−2/n ≤ a′n ≤ Nα−1/n;Nβ−2/n ≤ b′n ≤ Nβ−1/n;Nγ−2/n ≤ c′n ≤ pcNγ−1/n.Hence, it follows that for A,B,C as functions of N thatlog Rad(pAa a′n)log(pAa a′n + pBb b′n)∈[log(a′n)C log(pc) + log(c′n),log(pa) + log(a′n)(C − 1) log(pc) + log(c′n)]⊆[(α− 2/n) log(N)(1− γ + γ − 1/n) log(N) + log(pc) ,(α− 1/n) log(N) + log(pa)(1− γ + γ − 2/n) log(N)− log(pc)]=[α− 2/n1− 1/n+ log(pc)/ log(N) ,α− 1/n+ log(pa)/ log(N)1− 2/n− log(pc)/ log(N)],andlog Rad(pBb b′n)log(pAa b′n + pBb b′n)∈[log(b′n)C log(pc) + log(c′n),log(pb) + log(b′n)(C − 1) log(pc) + log(c′n)]⊆[(β − 2/n) log(N)(1− γ + γ − 1/n) log(N) + log(pc) ,(β − 1/n) log(N) + log(pb)(1− γ + γ − 2/n) log(N)− log(pc)]=[β − 2/n1− 1/n+ log(pc)/ log(N) ,β − 1/n+ log(pb)/ log(N)1− 2/n− log(pc)/ log(N)],andlog Rad(pAa a′n + pBb b′n)log(pAa a′n + pBb b′n)∈[log(c′n)C log(pc) + log(c′n),log(pc) + log(c′n)(C − 1) log(pc) + log(c′n)]⊆[(γ − 2/n) log(N)(1− γ + γ − 1/n) log(N) + log(pc) ,(γ − 1/n) log(N) + log(pc)(1− γ + γ − 2/n) log(N)− log(pc)]=[γ − 2/n1− 1/n+ log(pc)/ log(N) ,γ − 1/n+ log(pc)/ log(N)1− 2/n− log(pc)/ log(N)].As N ≥ n, it follows that,limn→∞log Rad(pAa a′n)log(pAa a′n + pBb b′n)= α,limn→∞log Rad(pBb b′n)log(pAa a′n + pBb b′n)= β,limn→∞log Rad(pAa a′n + pBb b′n)log(pAa a′n + pBb b′n)= γ.90Observe that since a′n ≥ Nα−2/n, it follows a′ ≥ nα−2/n. This implies that there are infinitelymany unique triples (a′n, b′n, c′n) derived from the triples in T . Recall from the beginning that forx, y ∈ N,−→iq(x, y) = (log x+ y)−1(log Radx, log Rad y, log Radx+ y),So we thus have that limn→∞−→iq(pAa a′n, pBb b′n) = (α, β, γ). This gives us (α, β, γ) ∈ Q′.91Chapter 7Conclusion7.1 Binary Forms, Polynomial Identities, and Limit Pointsin Three DimensionsOur work on polynomial and binary forms in terms of a three-dimensional vector of the inversequality reveals that the techniques used in the one-dimensional case are something akin to anoptical illusion. Results in terms of limit points in the one-dimensional case give closed lines thatcover almost all (or, when the abc conjecture is assumed, all) of the possible space. However,when one determines what these results mean in three dimensions, one obtains a set of disjoint linesegments and various points lying in an empty volume.This is not to say that the one-dimensional results and techniques are not impressive. Theystill give us the points with minimum inverse quality: (2/5, 19/60, 7/20), (19/60, 2/5, 7/20) ∈ Q′unconditionally. However, any attempt to translate or apply them to obtain a volume seems to fail.In fact, it seems that in order to obtain a volume with those techniques, it would be necessary tofind an infinite set of polynomials (or binary forms) whose identities could be manipulated in theproper manner. This is not necessarily impossible, but given the restrictions that apply to say, theBFGS Theorem, it is certainly no easy task.In many ways we are left with more questions than answers. Given the extent to which, withoutassuming the abc conjecture, it is possible to obtain points close to the boundary x+ y + z = 1, itseems possible that the BFGS Theorem, or an improvement of it, combined with a sufficiently largefamily of the right binary forms, could produce not only a volume but a large volume. However,it is not clear at all where one would start to look for such forms; many of the families discussedin [1] seem interesting at first, but the best application of them to the problem at hand is not at92all obvious.7.2 Circle Method, Possible Optimizations and Generaliza-tionsThe end result of our work in §6 was to prove Theorem 2.2. The volume described in this theorem,R ⊆ Q′, has the following geometric properties:• The point on the boundary of R nearest the plane α + β + γ = 1 is (7/9, 7/9, 7/9). Thiscorresponds to the inverse quality of 21/9. Polynomial methods get closer but this is theclosest the circle method comes.• The volume of R is positive.The volume R falls far short of the ideal volume of H. This was expected, given the difficultyof obtaining one-dimensional limit points with inverse quality near one. However, there are severalpossible avenues for improving the result. As the volume R is directly related to the size of thevarious error terms computed in §6, improving R is simply a matter of finding better bounds onthe various errors.The first source of errors come from the ∆∆∆ type integrals (see §6.5). In the current com-putation, these error bounds are strictly smaller than the error bounds for f?∆∆ (see §6.6) typeintegrals, so optimization of these should not be the highest priority. However, if these at somepoint need to be optimized, the most likely means of doing so is by finding a smaller bound on theL3-norm for the ∆ function from [2] over the major arcs, and a smaller bound on L3-norm for thef function over the minor arcs. Such improved bounds may be difficult to find for the L3-normdirectly, so it may be better to first compute the L4-norms for the ∆ function over the major arcs,and the L4-norms for the f function over the minor arcs.Similarly, one may look at the f?∆∆ type integrals. These integrals have the most substantialerror term and give rise to the pyramidal tip farthest from the trivial corner. Improving these errorswill have an immediate effect on the volume of R. Like the ∆∆∆ type integrals, optimizing theL3-norms of the ∆ function on the major arcs and the f function on the minor arcs will shrink allthe errors in this section. The L3-norm of f? on the major arcs cannot be optimized further by thesimple observation that the truncated exponential sum has an exact value, and that value is thegrowth rate used: it cannot be improved.There are also possible improvements in technique in the f?∆∆ section. Lemma 6.43 is partic-ularly bad, and it is possible a carefully constructed dyadic dissection would improve it. In general,in this section, it is likely that the dyadic dissections used could be improved upon. In particular,93finding the right parameters to allow the dissection of the minor arcs along with the major arcswould yield a substantial improvement. It should be noted that a na¨ıve attempt at this will fail, asthe bound of f along the minor arcs increases once the R parameter is sufficiently small; the propersetup must account for this.One final possible method to improve the f?∆∆ error terms is to somehow factor the notionthat the minor arc integrals are not over whole minor arcs but a smaller subset of distorted majorarc images lying in the minor arcs. No technique seems to work correctly here, but if it could bedone, it may substantially improve the error terms.The f?f?∆ type integrals (see §6.7 and §6.8 are fairly distinct. If one is content to remain in thecube [1/2, 1]3, it may suffice to improve the other errors and accept the limitations of the technique.However, if the L3-norms are sufficiently improved, and techniques for scaling the R parameterswith regard to minor arcs are refined, it may be possible to replace the atypical split used in f?f?∆with the three L3-norm Ho¨lder inequality used in the previous two sections. If this is done, andthe bounds are good enough, it may be possible to escape from this cube.Finally, regardless of how much these error terms improve, there is a hard limit on the volume oflimit points that is independent of R: the function U(N) described in §6. This function is essentialto the computation of the limit points, as it gives the growth rate for the main term. Recall thatits formula is U(N) = Nα+β+γ−1 + O(Nα + Nβ + Nγ). Thus the main term is greater than theerror term as long as α + β > 1, α + γ > 1, and β + γ > 1. Thus these three half-planes put anultimate bound on the error terms.Outside of improving the volume, one may also consider generalizations of the function F (N)described in §6. One easy generalization, which was not done solely for the sake of simplifyingcalculations, is to switch from three arbitrary distinct primes pa, pb, pc to instead three arbitrary,relatively prime squarefree natural numbers sa, sb, sc. There is no reason for this not to work, as atno point is the primality of pa, pb, pc invoked except to simplify some calculations. Going further,and replacing pa, pb, pc with arbitrary, relatively prime natural numbers na, nb, nc may also work.Finally, we shall note that the asymptotic formula for F (N) may be of interest in and of itself,as it counts the multiples of large powers of primes.7.3 Relationship with Kane’s PaperThe work we have done is similar in several respects to a paper by Kane [4]. His results are of asufficiently similar nature that a discussion the relation between his result and our result will aidin the understanding of both. We seek to show that both our result and Kane’s are, despite beingsimilar in form, sufficiently different that the task of showing whether or not they are equivalent94problems is non-trivial.Kane is concerned with the bounds of a function, Sα,β,γ(N):Definition 7.1. For α, β, γ ∈ [0, 1], let Sα,β,γ : N→ N be the functionSα,β,γ(N) = #{(a, b, c) ∈ Z3 : a+ b+ c = 0,Rad a ≤ |a|α,Rad b ≤ |b|β ,Rad c ≤ |c|γ ,max{|a|, |b|, |c|} ≤ N}.By comparison, our counting functions, described in §5 and §6 count subsets of N3 wherec = a + b, thus ensuring that c > max{a, b}. Kane’s function counts a set closed under anypermutation, while our functions count sets that only allow the permutation of a and b. Thusour counting functions are sensitive to asymmetries in the distribution of radical sizes betweenthe sum and the summands, while Kane’s is not. While there is no reason to expect there to beasymmetries of this sort, the absence of such an asymmetry would be desirable to prove, which iswhy we constructed our functions in this manner.Returning to Kane’s result, his paper was concerned with finding the largest set of (α, β, γ) ∈[0, 1]3 such that the following inequality held: that for all  > 0, and N sufficiently large,Nα+β+γ−1− < Sα,β,γ(N) < Nα+β+γ−1+.Kane’s exact results are slightly more precise than these bounds, but the set of (α, β, γ) ∈ [0, 1]3that Kane can prove have the desired bounds for any N sufficiently large is the regionK = {(α, β, γ) ∈ [0, 1]3 : α+ β + γ ≥ 2},with the lower bound being close for α+ β + γ > 1.It should be noted that our methods cannot improve on Kane’s bounds for Sα,β,γ . The functionthat we have that would be best suited to this would be U(N) from §6, which is as described onlysuitable for finding a lower bound to Sα,β,γ , and has error terms that dominate the main termbefore α+ β + γ > 1.Likewise, it should be noted that from the definition of the bounds above, obtaining limit pointsfrom the bounds of the counting function of Sα,β,γ is non-trivial: since Sα,β,γ counts a, b, c ∈ Z,where log Rad a ≤ α log |a|, log Rad b ≤ β log |b|, and log Rad c ≤ γ log |c|, the ranges of the sizes ofthe radicals are too broad to simply take elements of this form, take N to infinity, and construct asequence that gives a limit point.However, despite this obstacle, it should be noted that under reasonable assumptions about thedistribution of squarefree numbers, it would seem that this problem could be overcome. However,such assumptions do not constitute proof, as it is unclear how the squarefree numbers are distributedin the set Λ ∩ P .95Given the difficulties, above, the question arises: can one extract a limit point, not from theinequality, but from Kane’s proof of the lower bound? To do so, we shall consider the techniquesused by Kane to find his lower bound.To prove his lower bound for a given α, β, γ and N , Kane constructs a latticeΛ = {(a, b, c) : a+ b+ c = 0, 2xqw | a, 3y | b, 5z | c},where x, y, w, z, q ∈ N are chosen such thatN1−α ≤ 2xqw/Rad(2xqw) ≤ 2qN1−α,N1−β ≤ 3y/Rad(3y) ≤ 3N1−β ,N1−γ ≤ 5z/Rad(5z) ≤ 5N1−γ ,and that 5 < q  logN is prime and for any A,B,C ∈ Z where 2xqw | A, 3y | B and 5z | C,max(|A|, |B|, |C|) ≥ Ω(N1/2−α−β−γ/ log(N)).By means of a geometry of numbers argument involving the volume on the polygon P where|A| ≤ N, |B| ≤ N, |C| ≤ N , and Lemma 6.13, he proves that Sα,β,γ = Ω(Nα+β+γ−1 log(N)−2).However, this system cannot be used to find limit points in Q′. The issue lies in the fact that forany (a, b, c) ∈ Λ ∩ P , the vector inverse quality is of the form (assuming without loss of generalitythat |c| > |a|, |b|),−→iq(a, b) =(log Rad alog c,log Rad blog c,log Rad clog c).It is clear from the nature of the problem that there is no non-constant lower bound forlog Rad a, log Rad b and only the upper bound of log Rad a  α logN, log Rad b  β logN . Thiswide range effectively prevents the finding of a limit point, at least, not without substantial work.There are also some difficulties in reconciling the sizes of a, b, and c, but those could be overcomewith similar techniques as found in the geometry of numbers section of §6.Thus we can conclude that there is no simple logical bridge between Kane’s result and ours.However, it may be the case that there is a bridge of sorts. Kane uses a slightly different formulationof a lattice than we do in order to obtain a superior range of lower bounds for his problem. Inour treatment of the function U(N), we use a weaker but simpler construction of a lattice tocount possible solutions before filtering them to be squarefree. While our choice of lattice does notcurrently affect our set of limit points, as other error terms dominate long before the errors of U(N)are reached, should improvement to those bounds result in the errors of U(N) becoming limiting,reformulating the problem so that U(N) can use Kane’s more refined construction of a lattice mightbe a good idea, and would constitute a bridge between results, albeit a very non-trivial one.As for the reverse, going from our results to Kane’s result, since our result covers less threedimensional space than Kane’s, there is little motivation to explore it in detail, but the counting96function F described in §6 can also be applied to the problem in Kane’s paper. We shall sketcha quick, informal argument here, as the full details would amount to a nearly complete repetitionof the proof of our theorem. Let Fα,β,γ be this function with parameters α, β, γ. By the sameargument as in the proof for Theorem 2.2, one obtains a sequence an, bn such that an = pAa x,bn = pBb y, an + bn = pCc z, where for each an, bn, pAa ≈ N1−α, pBb ≈ N1−β , pCc ≈ N1−γ , andx, y, z ∈ N are squarefree and x ≈ Nα, y ≈ Nβ , and z ≈ Nγ . This would then give a lower boundon Sα,β,γ .97Bibliography[1] J. Browkin, M. Filaseta, G. Greaves, and A. Schinzel. Squarefree values of polynomials and theabc conjecture. In G. Greaves, G. Harman, and M. Huxley, editors, Sieve Methods, ExponentialSums, and their Applications in Number Theory, London Mathematical Society Lecture NotesSeries, pages 65–86. Cambridge: Cambridge University Press, 1997.[2] J. Brudern, A. Granville, A. Perelli, R.C. Vaughan, and T.D. Wooley. On the exponential sumover k-free numbers. Philosophical Transactions of the Royal Society of London A, Vol. 356,Issue 1738:739–761, 1998.[3] M. Filaseta and S. Konyagin. On a limit point associated with the abc conjecture. ColloquiumMathematicum, Volume 76, No 2:265–268, 1998.[4] Daniel M. Kane. On the number of abc solutions with restricted radical sizes. Journal of NumberTheory, Vol. 154:32–43, 2015.[5] D. I. Tolev. On the exponential sum with square-free numbers. Bull. London Math. Soc, Vol.37, Issue 6:827–834, 2005.98Appendix AIntersection of a Plane andRectangular PrismIn this appendix, we write the technical details involved in computing the volume and perimeter ofthe polygon PN described in Lemma 6.15Let N > 0, PN ⊆ R3 be the polygon produced by the intersection of the solid rectangular prismRN = (0, Nα]× (0, Nβ ]× (0, Nγ ] and the plane VN = {(x, y, z) ∈ R3 : pAa x+ pBb y = pCc z}.Let ρ =√1 +h2ch2aN2α−2γ + h2ch2bN2β−2γ .By the definition of PN , one may observe that its boundary is the closed loop contained in theintersection of the planes defining the faces of RN and the planar vector space VN , which is thefollowing set of lines:`x1(t) = (1, 0,pAapCc)t, `x2(t) = (1, 0,pAapCc)t+ (0, 1,pBbpCc)Nβ ,`y1(t) = (0, 1,pBbpCc)t, `y2(t) = (0, 1,pBbpCc)t+ (1, 0,pAapCc)Nα,`z1(t) = (1,−pAapBb, 0)t, `z2(t) = (1,−pAapBb, 0)t+ (0,pCcpBb, 1)Nγ .Note that lines with the same letter in the subscript are distinct parallel lines as N > 0.Thus the vertices of PN must be a subset of the intersections of the lines given above. Excludingrepeated vertices and those that unconditionally lie outside the first octant, the following points99become the possible vertices of PN in addition to the origin:piX1 = `x1 ∩ `y2 = (Nα, 0, hchaNγ),piX2 = `x1 ∩ `z2 = (hahcNα, 0, Nγ),piY 1 = `x2 ∩ `y1 = (0, Nβ , hchbNγ),piY 2 = `y1 ∩ `z2 = (0, hbhcNβ , Nγ),piZ1 = `x2 ∩ `z2 = ((hahc− hahb)Nα, Nβ , Nγ),piZ2 = `y2 ∩ `z2 = (Nα, (hbhc− hbha)Nβ , Nγ),pi1 = `x2 ∩ `y2 = (Nα, Nβ , (hcha+hchb)Nγ).To complete the proof, start with the fact that (0, 0, 0) is a vertex of PN and `x1 a line intersectingan edge of the boundary of PN . By traversal of this line in the direction (1, 0, 0) (since PN liesin the first octant) until the first intersection of two lines on the boundary is reached, rotatingthe direction of traversal counterclockwise to match the new line, and repeating the process untilarriving back at (0, 0, 0) it is possible to trace the polygon PN .Triangle. Suppose ha ≤ hc and hb ≤ hc.Observe that traversal starts at (0, 0, 0), goes along `x1 until the intersection piX2 which occursbefore piX1 since ha/hc ≤ 1. Now traversal occurs on `z2 until the intersection piY 2 whichoccurs before piZ1 since hb/hc ≤ 1. Finally, traversal moves down along `y1 and returns to(0, 0, 0).pi0 → piX2 → piY 2 → pi0.Thus PN is a triangle andPerimeter(PN ) = d(pi0, piX2) + d(piX2, piY 2) + d(piY 2, pi0).The projection of PN onto the xy plane SN has vertices(0, 0), (hahcNα, 0), (0,hbhcNβ)and thus by the formula for the area of a triangle,Volume(SN ) =12hahbh2cNα+β .100Quadrilateral I. Suppose ha ≤ hc and hb > hc.Observe that traversal starts at (0, 0, 0) along `x1 until piX2, same as in the previous case.Now traversal occurs along `z2 until the intersection at piZ1 which occurs before piY 2 sincehb/hc > 1. Now traversal moves along `x2 towards the y-axis until the intersection at piY 1.As hb/hc > 1, there are no intersections between piY 1 and the origin and so the traversalcompletes the circuit.pi0 → piX2 → piZ1 → piY 1 → pi0.Thus PN is a quadrilateral andPerimeter(PN ) = d(pi0, piX2) + d(piX2, piZ1) + d(piZ1, piY 1) + d(piY 1, pi0).The projection of PN onto the xy plane SN has vertices(0, 0), (hahcNα, 0), β), ((hahc− hahb)Nα, Nβ)(0, Nβ)and thus by representing the area of SN as the subtraction of the area of a triangle from arectangle,Volume(SN ) =hahcNα+β − 12hahbNα+β .Quadrilateral II. Suppose ha > hc and hb ≤ hc.Observe that traversal starts at (0, 0, 0) along `x1 until reaching piX1, which occurs beforepiX2 since ha/hc > 1. Traversal continues along `y2 until the intersection at piZ2 which occursbefore pi1 since hb ≤ hc and lies in the first octant because ha > hc. From piZ2 traversalcontinues along `z2 until piY 2 which occurs before piZ1 since hb/hc ≤ 1. From there traversaldescends back to (0, 0, 0).pi0 → piX1 → piZ2 → piY 2 → pi0.Thus PN is a quadrilateral andPerimeter(PN ) = d(pi0, piX1) + d(piX1, piZ2) + d(piZ2, piY 2) + d(piY 2, pi0).The projection of PN onto the xy plane SN has vertices(0, 0), (Nα, 0), (Nα, (hbhc− hbha)Nβ), (0,hbhcNβ)and thus by representing the area of SN as the subtraction of the area of a triangle from arectangle,Volume(SN ) =hbhcNα+β − 12hbhaNα+β .101Parallelogram. Suppose ha > hc, hb > hc andhahc− hahb ≥ 1.First note thathb/hc − hb/ha = hb(1/hc − 1/ha) = hb(ha/hc − 1)/ha ≥ hb(ha/hb)/ha = 1.Observe that traversal starts at (0, 0, 0) along `x1 until piX1 as before. Traversal continuesalong `y2 until pi1 since hb/hc−hb/ha ≥ 1 so pi1 occurs before piZ2. From pi1 traversal continuesalong `x2 until piY 1 as by assumption the x coordinate of piZ1 is larger than Nα. From piY 1the traversal returns to the origin along `y1.pi0 → piX1 → pi1 → piY 1 → pi0.Thus PN is a parallelogram andPerimeter(PN ) = d(pi0, piX1) + d(piX1, pi1) + d(pi1, piY 1) + d(piY 1, pi0).The projection of PN onto the xy plane SN has vertices(0, 0), (Nα, 0), (Nα, Nβ), (0, Nβ)and thus by the formula for the area of a rectangle,Volume(SN ) = Nα+β .Pentagon. Suppose ha > hc, hb > hc andhahc− hahb < 1.First note thathb/hc − hb/ha = hb(1/hc − 1/ha) = hb(ha/hc − 1)/ha < hb(ha/hb)/ha = 1.Observe that traversal starts at (0, 0, 0) along `x1 until piX1 as before. Traversal then goesalong `y2 until piZ2 since hb/hc − hb/ha < 1 so piZ2 occurs before pi1. From piZ2 traversalcontinues along `z2 to piZ1 which occurs before piX2 since hb/hc > 1. From piZ1 along `x2traversal goes to the next intersection at piY 1, then descends to along `y1 to the origin.pi0 → piX1 → piZ2 → piZ1 → piY 1 → pi0.Thus PN is a pentagon andPerimeter(PN ) = d(pi0, piX1) + d(piX1, piZ2) + d(piZ2, piZ1) + d(piZ1, piY 1) + d(piY 1, pi0).102Thus the projection of PN onto the xy plane SN has vertices(0, 0), (Nα, 0), (Nα, (hbhc− hbha)Nβ), ((hahc− hahb)Nα, Nβ), (0, Nβ)and thus by subtracting the upper right triangle from the rectangle, one has thatVolume(SN ) = Nα+β − 12(1− hahc+hahb)(1− hbhc+hbha)Nα+β ,= (1− hb2ha(1− hahc+hahb)2)Nα+β .It is clear from the inequalities that this covers every possibility.The determination of the area of PN as opposed to SN is done by application of Lemma 6.14.The lower bounds on the coefficients of the area are determined by observing that quadrilateralI contains the triangle pi0 → piX2 → piY 1 → pi0, quadrilateral II contains the triangle pi0 → piX1 →piY 2 → pi0, the parallelogram and pentagon contain the triangle pi0 → piX1 → piY 1 → pi0, so itfollows that the area of the polygon is minimal implies that the polygon is a triangle. Then it is asimple observation that as 1 ≤ ha < pa, 1 ≤ hb < pb, 1 ≤ hc < pc,hahb/(2h2c)ρNα+β ≥ 1/(2p2c)ρNα+β .The upper bound is determined by observing that all five polygons lie in the parallelogrampi0 → piX1 → pi1 → piY 1 → pi0, so the maximum area of PN is thus ρNα + β.We now move on to the perimeter. We calculate the distance of each line segment.• pi0 → piX1 has length√N2α +h2ch2aN2γ .• pi0 → piX2 has length√h2ah2cN2α +N2γ .• pi0 → piY 1 has length√N2β +h2ch2bN2γ .• pi0 → piY 2 has length√h2bh2cN2β +N2γ .• piX1 → piZ2 has length√(hbhc− hbha)2N2β +(1− hcha)2N2γ .• piX1 → pi1 has length√N2β +h2ch2bN2γ .• piX2 → piZ1 has length√h2ah2bN2α +N2β .• piX2 → piY 2 has length√h2ah2cN2α +h2bh2cN2β .103• piY 1 → piZ1 has length√(hahc− hahb)2N2α +(1− hchb)2N2γ .• piY 1 → pi1 has length√N2α +h2ch2aN2γ .• piY 2 → piZ2 has length√N2α +h2bh2aN2β .• piZ1 → piZ2 has length√(1− hahc + hahb)2N2α +(1− hbhc + hbha)2N2β .The perimeters given in the lemma are simply sums of these.104

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