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Determination of bases for certain quartic number fields Murdock, David Carruthers 1933

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. DETEP-MIE&TIOII OF BASES FOR CERIAIli QEARTIG ,-IUlCBER FIELDS .David. Carru.th.ers Murdoch A Thesis submitted f o r the Degree of MASTER OS ARTS i n the Department of IKE UITIYERSIIY OF BRITISH COLUMBIA May, 1933. m_HlMATICS Determination of Bases for C e r t a i n Quartic lumber F i e l d s . Introduction. This thesis deals with the determination of bases f o r integers of the quartic number f i e l d defined by the equation ^ + = ° • The method used i s that developed by F. R. Wilson i n his paper "Integers and Basis 1 • • of a Number F i e l d " . The notation i s , i n the main, adopted from Wilson»s paper. The l e t t e r s and ^ and X , Y, and 2 denote algebraic integers, while the remaining l e t t e r s , Q-, -fr -c, w and t h e i r c a p i t a l s represent r a t i o n a l i n t e g e r s , ^ specifying a prims. Greek l e t t e r s denote r a t i o n a l numbers. The notation a used for a . divides and (X f o r CL does not divide We now give a summary of the p r i n c i p a l definitions; and theorems of Wilson's paper* These are stated for the general f i e l d of the n**1 degree defined by the equation yL + 3^.fyc + - + J30 — 0 This- equation, i s assumed to be i n the normal form, that i s 30 ,3,, are a l l r a t i o n a l integers and there exists no prime such that (f^/-B^-si. f o r ®.ll of Sl~ -1,2-—/"?\. 1. Transactions of the American Mathematical Society, Vol. 29, 1927. I t i s obtained from the general equation . --M..= <? by the transformation d-n = ^ A^'j.-f-/)^., where oL i s the greatest r a t i o n a l integer such that the r e s u l t i n g c o e f f i c i e n t s , 3o, 3,>~ ~ > 3*T~--/ are a l l r a t i o n a l integers. The integer a I 3 ^ , ^ , • ^ and a l l l i n e a r combinations of them with r a t i o n a l i n t e g r a l c o e f f i c i e n t s are. c a l l e d ordinary integers. An integer of the form t = Otf0 +• ptyc + — -f- (X^tC , symbolized by (_<*o j , } > CK^) ; i n which the denominators of a l l the or are powers of a single prime, and ^ O , i s c a l l e d a single-prime integer of degree ^ in yc . I f , moreover, or^ r _n > 0, ajad i f then -^ i s c a l l e d a single-prime reduced integer. I f jf i s a single-prime reduced integer and i f c/^. = -yV where t i s the greatest possible, then ^ i s c a l l e d a maximal  reduced integer of degree >•?»_ i n . The theorems of Wilson 1 s paper are as follows; Theorem I. I f £ H i s the highest power of Jz occurring i n the denominator of any a of a single-prime reduced integer then p i s a factor of the discriminant of the •field, equation. Corollary 1. I f there exists a single-prime reduced integer i n ^ of degree i n -x } there also e x i s t s a maximal reduced integer i n ^ of degree i n -X » C o r o l l a r y 2. The maximal reduced integers i n a given f i e l d are f i n i t e i n number. Theorem I l ( a ) . A l l integers of the f i e l d can be expressed as r a t i o n a l l i n e a r homogeneous functions with r a t i o n a l , i n -t e g r a l c o e f f i c i e n t s , of ordinary integers and maximal r e -duced integers, the l a t t e r consisting of ons selected a r b i t r a r i l y from those i n each prime and for each degree i n •x. for which such e x i s t . C o r o l l a r y . The difference between two integers i n ^> of the same degree i n yc and having the same highest c o e f f i c i e n t i s expressible i n terms of ordinary and maximal reduced i n -tegers of lower degree i n ^  . Theorem I l ( b ) . I f 6^ i s a prime occasing i n the denominator of some co-ordinate of an integer, then ( l ) there e x i s t s exactly one maximal reduced integer, Ysl i n ft of lowest degree -si i n ix- } st >0; (2) i f ~^ -"*"--/ there e x i s t maximal reduced integers of degrees-^T-/,-A./* 1} ^ >*\-x and (3), for each u. } o < V- <£. t } there is one and but one single-prime reduced integer ^/30>/3,) J o f i e g r e e si- in. -vc d i f f e r i n g from £ ' by ordinary integers, where YX = ' fro > Y,, ^ r j Theorem I l ( c ) . I f l^j , ^  , -~~h^} i s a maximal reduced integer i n >^ of degree . > w ; i n -?c } then £4 =^ f o r .<s«c-^»^ Now i f i s the lowest degree i n -x. f o r which a maximal reduced integer occurs, f o r any K , st H _= -*<--/, we l e t be any se l e c t i o n of maximal reduced integers of degree K i n -T^T one for each d i s t i n c t prime for which such occur. We also l e t 7^ ..= ^ *• .. and l e t ^ , ^  , ---be non-zero solutions of Then i f 2 i s *^e' reduced integer derived from y- i/ f (j- (, + by removing ordinary integers so that •_ ' < (y . __ __• • • • then the basis of the f i e l d i s given by It only remains to determine the complete set of maximal reduced integers (fk't '•" ^ i s Wilson proves the following theorems. Theorem III» I f A, and A denote the discriminants of the f i e l d and of the equation defining the f i e l d , r e-spectively, and i f Tli , 7°*+, } "75!-/ are the denominators of the highest c o e f f i c i e n t s of the elements of the basis as above determined, other than those which are ordinary integers, then A = f^B^r- - £ w ^ / Theorem IV (a) « It p. = +• y* + -t of^_l yc is: an integer and i f El i s the eliminant of and the f i e l d equation, then a l l of and a l l of - / L - I j X , - - --;^-(, are r a t i o n a l integers; and conversely, i f a l l of — V > ^ = 0 , 1 , 1 , are r a t i o n a l integers, then ^. i s an integer. Theorem I V & ) For the f i e l d ^ ^ - ^ V - r +B0 — 0 the maximal reduced integer ' jpO can exis t only i f (1) J>*^~0t/U ; (2) ^ V ; ff%> • y^'and.(37 either <2.0 =. O c^l -ft l^r\.. Theorem V . When a maximal reduced integer i n ^ of degree ~-*t-/ i n -?c e x i s t s hut none i n /> of lower degree i n -pc there exists a single prime reduced integer such that either ( l ) , jj-x reduees to ordinary integers and a _ 3 ~ B ^ . x > « 0 = 0==B* , a l l mod ^  ; or ( 2 ) , £ X does not reduce to ordinary integers and 3 —->t-__ Corollary • 1. I f ^ i s a f a c t o r of each of 3 0 ,3, , 3^_t and a maximal reduced integer i n A of degree >^->~f i n exists "but none of lower degree i n yc then ^ — i s an Integer* Corollary 2. I f ^ i s a factor of 3OJ^/} ^ but not of 3sx+t and there i s a maximal reduced integer of degree ^v-/ i n ~pc but none of lower degree i n then £0 - a, = = = 0 and ( l ) i f CL^_^ s B^-, mod^ , -Qjx^ 0 or (.2) i f a^_-x 4 3^., mod^ , then and CL* +, 7^ 0 • C o r o l l a r y 3 « I f ^  i s a factor of S^.-, hut not of _ 5 0 then a6¥ O and ^ O . - 6 -2. The Eliminant and Discriminant of the F i e l d . The eliminant of the equation <- X*. +S = o and the general element, . .y = <*_ r «3 -x f of the corresponding f i e l d , i s given Dy the norm of .^, that i s by the product of the four conjugates •x 3 1 J i oC0 ex, -X3 +• « A -Xz .+ -efyTj in i 3 where , , *x3 , and ^ v are the roots of 7C -f- Qrx. -t- H-x + S - Q This product i s C EE <*o -t- *o <x, e*0 c^Zx, +• «*0 of^ Z ^c, + or0 of, Z rx, ' »• < 1 _ * 3 3 3/ 3 / I t * . * """a <*_. * Z ycx + <*t «, 2 x, * t ^ 5 -f- <x0o<K Z T<x0-ex3 Z. ?cr zljl-?t.3 s- ex, -Xf-x^-xs4*^ + <X, of^Z. -x, ^ x^y . / J 3 3 2. ~*3 "X-tf "f *o Z X , ^3 +• °<o <*, <* 3 Z •X.,'XL. Hs + *» <*1 Z -K, ~X-K ^ « r 3 Z -h <*o *% o<3 Z yCf^X ^3 -The c o e f f i c i e n t s of the <x are symmetric poly-nomials i n , -x^ , ^  , and and. can therefore be ex pressed, i n terms of the elementary symmetric functions, £•*,., £ -xx , £ ^ ^ - ^ j , and. : -x h and hence i n terms of Ql , 7?, and S. The results are a follows o - 1Q - 7? £ ^_ y * 5 £ * , - 3 n a Z 3 3 3 = s £. ^ - O * 5 ~^ H 41 i. i_ Z rc^ -7C3 = -KS.S 4 i 3 Z x A ^ j - x ^ = Q S +%S = - * s £ */ ?«j * v = - TPS / J J 3 Z ^ t *^ — = s ••/••• 4- 4. Z - .x , * 4 •vc2 - a n z a *• t. • s. 2- 3 c 4 ^ 3 = 0 . - =r - h S Z 3 3 t L Z ^ 1 *j = a s x ^ v x i Z. - x , •x'j — . y e s / J J 3 X 2_ ^ 7C3 Kv Z . * 1 2L -X* -*3 — 7c, ?<x -x^ y<¥ 3 =• S , 4 3 3 — -£ ^/ 2, "PS-i X 3 Substituting these i n the expression f o r £ we get E •==•• «V - i-Q. tx/cx^ — 3 %*oc<y I-Quocx? + (Qx+is)<x**l +-($/ ? 3 R X - 3 Q S ) t * / - ft cx0 „/ + (R1-- ±&s)cx0<y/ + (tQKS-tf)*, <y3 + 5 <*, - L&S cx, or3 •+ Q.S cx, cx% +{GlS +S-S JcX,oc3 -US <x,cxx 3 j/ - 9 -(hQl S - 3 .* *J *<> • _fow for the f i e l d * <3 ^  * s - o we have 7?=o and the eliminant reduces to 3 -^ 6 o', — 1 AS<*V cx3 f- QS <*/ cxu +• (QS't-Sj ~ l&S c<, cx^ + S**t-•+ •Q.s''txlot3-- + S 3cx3 +.(us •-%<£) ce?.oe, #2 MS«,*,*••«± - //glo</oc^a3 / (u S^- J- &XS) <x0 t tf fiS *0 of, *% *3 . This r e s u l t has been checked by Sylvester's method of elimination. We also have "a 2-\ dcx^dcx, <y, <x + &% -x- +• off ? By Theorem 17(a) (introduction), i f = #o + i s an integer then a l l the above expressions are r a t i o n a l integers, and, conversely, i f ^ 77 - ^ r ^ = 0> 1 > % ) 3 , are r a t i o n a l integers then c^- i s an integer. To f i n d the discriminant of the f i e l d equation 7c + ft ^  +• S — o we use the fact that the discriminan" of any quartie i s equal to that of i t s resolvent cubic. In t h i s case the resolvent cubic i s ~ Q. \t *" - h S <t -/-uGiSeZkO or 7':-•*>/. The three roots are Q , Vms , and —Yus and hence the discriminant, A , i s given by o r - • - 11 -2. ReQapitulation« I f J[ ~ cs0 +. of, -j* 4- «^ + cx3 -pc i s an integer of the f i e l d . 3 6 *• 4 s = o , then the following are a l l r a t i o n a l integers. ly = - i. cx_ +M*etx, +(0- tzS/iXo «_ •+\@.--3GLSJ<x0 oc3 - ,2 6 S «•„ <<,-£. a$ <x, p<3 + Q.S, <x, «x +(Q $ +1$ Jo., or} - SlCS of, ^ 3 y. 5 «*_ 3 \^=. k 6 c l * 0 \ x + £&<xca>*> L(aLt-%s)x0(a3-3qs)#«<*3-ias J ' =_ - ^ficf. ^ _ * c t5J #i +. (a3- 2 as) ^ I 5fe)=aL f i - 5 ' f i s ; ^ 3 * ( * . s - a J 4 <^ .^A', Furthermore It I, , 1 1 , I3 t and 7^  are r a t i o n a l integers, then i s an integer of the f i e l d . The discriminant of the field, equation i s From theorem I, i f ^? i s the highest power of occurring i n the denominator of an integer, then £ (<a . The work of determining the maximal reduced Integers of the f i e l d i s divided into f i v e cases as follows: Case I; To determine the maximal reduced integer^of the 1st degree i n Case I I ; To determine the maximal reduced integers of the 2nd degree i n X when none of the 1st degree e x i s t . Case I I I ; To determine the maximal reduced integers of the 3rd degree i n when none of lower degree e x i s t . Case IV; To determine the maximal reduced integers of the 3rd degree i n x when those of the 2nd degree ex i s t but that of the 1st degree does not. Case V; To determine the maximal reduced integers of the 2nd and 3rd degrees i n Y whenone of the 1st degree e x i s t s . - 13 -3» Case I* I f there i s a maximal reduced integer of the 1st degree i n " i t i s of the form £ = (~^T<, '\ We cannot have aD ~ o for i f so, y = ^t, ^ and on sub s t i t u t i n g t h i s i n 13 and respectively, we get ^ 10. and 'jS which i s impossible since the f i e l d equation i s i n the normal form (see introduction). Moreover, t t o f o r i f t, > t0 then ^. leads to a single-prime reduced integer for which a.0 —O and t h i s has been shown to be impossible. By Theorem I l ( c ) ~ £ 0 and hence we must have t, — tb From Theorem I V ^ ) , since a0 o , we have £ j % and ^>tf/zQ. whence ^ = X and t0 _<__ z. and <j, reduces to (-JT > f ) where t £ i _ I f t = X on s u b s t i t u t i n g i n 1$ we get Q == - 6 mod, ^  , -b-at ftbj.3 i^--^g^s^ih±e^-^iTr(5ei—fre-m a-bove, —i:y^_K Therefore ^ ?=/ and since ~ X ^ X ^ and fi0 We have Gt> - / and ^ becomes ^  ; j ) Substituting /this i n JT, --- ?.-- — F H i s s a t i s -f ied identic/ally, i " ^ and _Tj i f Q. = 4., mod ^ and I, i : * S =E —(& v-/),mod /6 . I f these two conditions are sat isf ied, therefore, we have the maximal reduced integer / + ~* o 4 , Case II» We f i r s t prove the following: Theorem A. I f a maximal reduced integer i n ^ of degree X i n exists but none of lower degree i n then there exists a single-prime reduced integer jz~(aO}G-,,0 such that Since there exists a maximal reduced integer i n ^ > of degree X i n , there exists a single-prime reduced integer of the form ^ = ^(^° > at j 0- Since ^ i s an integer, so also i s jf-x - Reducing t h i s by means of the f i e l d equation we have The difference i s also an integer. The difference between a ' jf ~~ J^ and Caii'~ a° +Q)JJ- i s an integer of the 1 s t degree i n rxr and must therefore reduce to ordinary integers since, by the hypothesis of Case I I , there i s no single-prime re-duced integer of the 1s t degree i n -x. From th i s fact the congruences • follow at once. Y/e now divide Case II into four sub-cases;, A* and f/S B . and G» but _Q« / / a but-; Sub-case At- //& and 4/S » I f a maximal reduced integer e x i s t s , by Theorem A there i s a single-prime reduced integer ^ x (&o > Q-, > l) such that or, s i n c e ^ > / & a n d /S, •6, (a/^CLi,)^ a0a;-J. ^ From ( I ) , either £ 0 = O or =_ £<,,moa.^ or both. In any case (2) gives £ e = a, —0. Hence i f a maximal reduced integer exists, there exists a single-prime reduced integer ~7~ •• ' Substituting t h i s i n I, we get d> •• and 7j_ ., 7 0 , and 1H are s a t i s f i e d by the same condition. — 16 «•• Hence i f and. £IS there i s an integer ' i f ^ i s not maximal, "by Theorem 11(h), there must "be an integer of the form 1 ~# 1 ^ ) • cannot have "both a0 = 0 and <x,~o f o r i n t h i s case I, gives /__> and I3 gives fiX/Q. which i s im-possible since the f i e l d equation i s i n the normal form. _Jor can we have dc — 0 and &, ¥ O unless ^ =-5. f o r , i f ^ 9^  £ It} gives t/' Q- whence from T3 and Ii, we have •jfi*/9 . Hence i f ^ ^ i . the only possible integers of t h i s form are mod Substituting ^ i n -Z"^  we get i £ 0 ^ ? = £ ^ y^ 1 or £6.0fi= Gl + y&'f~. Substituting t h i s i n Is and we f i n d Q.r- h S = Q) mod^ " and Q.*- 4-S = £a.,y£\ mod^^ whence Q. and ^¥/S which i s impossible. There can, therefore, be no integer of the form Hence, i f the only possible integer of the 2nd degree i n jf? i s (~ } o } ^y,^ I f t h i s i s not maximal by Theorem I l ( b ) there i s an integer of the f orm ~p J .q ^ o where t < -^-f and ^ _= 3. We s h a l l show that t h i s cannot exist unless 0-, = O -- 17 -From Iu we have ± &eft = ^ m o i ^ ' ^ and hence, since a0 ^0 ' From T^Ca) , however, we get, assuming _2, ^  o / that ^ /d and therefore t-/ and our integer becomes 3 } From ./"j on s u b s t i t u t i n g la0jf> = Q. + , w e get Q - U S = Oj mod ^ 6 and on wr i t i n g 4S - f> and £2 —^ko? we f i n d i . a f ' a ^ ® We f i n d also that -Z* i s s a t i s f i e d by the same condition. From Xr when we put S- -fi^s' and #5 = £L mod ^  , we get But i n view of (3) thi s i s impossible for &, ^  O. Therefore a, = o and the only possible integer i s Substituting t h i s i n IH we g e t i . A o ^ = = £1 mod ^  and j T 3 , .2\ ; and I,give the congruences: 3 ,3 • , „ i /1 , _ i , j - 18 -S u b s t i t u t i n g xaBf> = fl ^ -f>™4 we f i n d that a l l three are s a t i s f i e d by the condition GL- H S= O, mod ^ . Hence we have the maximal reduced integer \^-i ' > ^J where --n. i s the greatest integer such that Q -:#Ss mod ^ , __ 1 , and / ^ c ^ 6 = a, mod ^ . We now consider the case ^ = 2_ . I f i s not maximal then there i s an integer of the form ( f f •> ~ ? - p Q It was shown above that we cannot have both &0- o and There are, therefore, three p o s s i b i l i t i e s (a) cio - o and (X, ^  o (b) a0 ^ o and a, ^ o and (c) a0 and a., •=. o . (a) Qlq- o and a, ^ o . Since ™ _ ^ x _= x w e bave CL.stl and the integer i s of the form (o, 4r , -k) This _/ cannot e x i s t , however, for Jf g i v e s ^ ^ r * yi aaad-—£ / *"> and hence requires that This i s impossible since the equation i s i n the normal form. (b) a0 *f o and ^ o , _. In this case a 0 =• / and -/ and the integer becomes (j. ' T ' "I1 } Substituting t h i s i n Z, — _T ¥ , we f i n d that Z y i s s a t i s f i e d , _T3 and Ix are s a t i s f i e d i f and S =_ 4> mod while I, requires that 1 /ft and S" == mod /6. I f these l a t t e r conditions are s a t i s f i e d , therefore^, there i s an integer > -£ > , - 1? -i f t h i s i s not maximal there i s an integer {~£ > -75 ' 2? i - This i s the only possible form, for \j} 3 "T ' i v would, lead to an integer (Ji y 0 > ~^-J which, subtracted from (•£ > _ 1 ~b-) , gives an integer ^ , which cannot ex i s t , while (•£ > > would, lead to the integer (0, ~%_ > which was discarded i n (a) • In con-f O-o CL, I \ .; ' s i d e r i n g ^ = ("J» > ~*f }~p) we s t i l l have the conditions 1*1 Q. and 5 = , mod /6} since the existence of ^ implies the existence of (%. 3 > i O We also have 0-0 = and <2, == -/- Substituting i n I,, IM ? XH i s s a t i s f i e d . Writing A = ?a' and S = /6-4 + H we f i n d that _~j i s s a t i s f i e d provided -^e i s odd. I f 0.0—j} Ix requires that 6>V-^= _),mod , and I, requires that G i ' V O mod. /6 ; but t h i s i s impossible, fo r from X 3 , i s odd and hence / mod 7/; g i v i n g •Q;'— / mod # which i s Impossible. I f a 0 - - / , _T, gives = mod /+ } and. 1, gives^p/f-/) t s c mod /^ -./y, which i s also impossible since -4. i s odd. Hence the i n t e g e r > > cannot e x i s t , and therefore we have the maximal reduced integer (J^ > >if)J the conditions for i t s existence being Q. = Otmod. 8" and 5 = ••tfi mod /_,... (c) a0 ^0 and &, —Q 9 I f there i s an integer of the form > o } "J-^J we must have ->-» > t as i t was shown above that (-£ > o, -jy.) cannot e x i s t . Substituting i n X „ , — ~r,I, , we get if. O-c S ' Q_.-^ywc£ (I) X H- Q-o ~ /*.*<>&+Gl + s-S = O ^ ^ d tL (Z) 3 i n + « & c a \ n e $ - ^ s = 0 ' ^ * ( X " - " O ) 76 ^"-./SaU+^a+ta's - H a 0 a s , S x = O ^ ^ U . --(*) From (I) we have - Q + £~ 'a . Substituting t h i s i n (2) we get ja. 'X — Q. + hS = Oswvd X and. therefore We now l e t US - Q.X + i * * %4- , ana su b s t i t u t i n g we get 30.^^4- = Oyyuuttl 8 (S) S i m i l a r l y , substituting i n (3) we get and. from (4) 4. == Oyy^K> ^ From (3) we have 4- - $•<• - 3&*'. Substituting t h i s i n (6) we get A<-a.'*_ 0 mods', and therefore X /0. and i f a. = % of we have •<2/4_= ^c, mod 2-- 21 -Hence, a.' and -e are either both odd or both even. I f aJ and -c are even, we have a- = o, mod and £a*V-^ ___> mod , and therefore = 0, mod /6 , and and ft^**^^ But i f a' and are odd we have from (i>) / 5- a,'1 -e- -6- = _) s^yMn&( % } ~£ 0 siwfv&L /6 and therefore Let -<£ = A-^ where i s odd, then and therefore since a. = / mod >*/ , we have 4- =.-j mod # -Therefore i f q\ us = l*(n<£-0 j •& there i s a maximal reduced integer 5 °> "^J where ^ = and ao == Q., mod 2. but =j£ Q., mod 5- . I f G -#S i s not of t h i s form, then i s the greatest integer such that . it--*, f a. /• > JL /(Q. -n-s) and Q.,mod£. . Sub-case B: ,£ j>& and,^ ^ jy. From Theorem A, i f there i s a maximal reduced integer i n ^ of degree. • X- i n s*r , there i s a single-prime reduced integer ^ =j(cLo , a,} /) such that a0 ( a,x-&o *-«.) = S J ^v^rd f> -~-Q) V 2- ^  t ' / . y / 2-- 22 -•• •CL;(CL,i'V G.0+-QJ == aoa, 1'. From l^ -fc,) o} mod 6^ , and. h e n c e s i n c e CL, — 0 Therefore, from ( l ) , since we must have ac^O and the only possible integer i s of the form ( "pr ' 1 ft J where O-^^O. A maximal reduced integer w i l l therefore be of the form ^ — ' 0 J •/ •£/ being zero from 3-s@-) • Consider,f i r s t ; the case ^ %. . Then from IH we get ^ a 0 = mod ^ or £ £ 0 — ft •+ ft 4i . Substituting t h i s i n X3 }'•• - T ' and JT, , we f i n d that they are a l l s a t i s f i e d by the condition Gl~-u-S =. o , mod ^ Therefore i f ^> ^ 1. there i s a maximal reduced integer > o'} -^J where ^ i s the greatest t'' •••• A^n Integer such that O- - US = o , mod ^ ^ and Xa0=. Q mod ^ Fc = the only possible integer i s / o f the form ^ ~ } - Substituting i n XHy^e get This can only be s a t i s f i e d i f pos s i b l e integer i s (-^ > O , get which i s impossible no maximal reuuced integer i f - 23 -Sub-case G; £ Is but ^ o^Q. * Again, from Theorem A, i f there i s a maximal reduced integer i n jf> of the 2nd degree i n nc then there i s a single-prime reduced integer ^ — ^C^o, j 0 such that i Q-c C°- #l - s S , W ^ (7J Since ^ / S ( l ) becomes CLo(o-r t-O-o •+-Q.) = D ^ iwd ^ ' (X-) From -/-fa.) ; = 0, mod ^  and therefore, since Q. we must have a,-0 . Hence from ( 2 ) , d0(a-^0)= 0,mo&^ and therefore either a c - 0 } or <2 0 = ^ ;mod We cannot have a0 — O f o r i n this case the integer becomes and su b s t i t u t i n g i n I3 we get . Q -f- X S ~ 0 sww&t which i s impossible since (f> / S but $ $ & The only p o s s i b i l i t y l e f t , therefore, i s ( ~ * ° J where ft-o s dy mod jft From X v , //fl0- o,mod ^  and therefore XGl=. O , mod /? / / ,: / <^<? <<-/ <— ^ . _.' • ^ ?iu-c*~*-/ / - 24 -Sub-case D; / g but & j>£ In t h i s case, i f there i s a maximal reduced, integer of the 2nd degree i n -?c then there i s a single-prime reduced integer —jr(&o » °-t •> 0 and since (ft/Q. the con-gruences of Theorem A reduce to a.a(&,1 — ao) ~. S <m^/^> ------------ (j) a, .(a;3" — #o) = a0ay^ym^/^ ---~ -• (%•) From (l) 7we cannot have a.0-o since jj£. The only pos s i b l e cases are therefore (a) cx0 ^  0 , CL, O and (b) a0 =£0 a, — O . Also from Theorem I, /£S (Q and therefore, since j,(a but rf>Js we must have ^ = jL • Hence the only possible integers are r = > •£ ' f_) and Substituting ^, we get, from J3 - -whence Q = x(i-s)^^/h-and therefore since S i s odd we have k /Q. , From Ix ft. or k ~ H&. + == Oyyuod% But t h i s i s impossible since and hence there can be no integer (jx ' I ' x) . We now substitute u i n _T, , — Ju . I I. I. are s a t i s f i e d while J., gives / — xq +•0*'f % S - zas + s r — 0,^uxl 16 or (7 -^1^ 3 O y^ad /6 whence -5 = - 0 *^u®c( [f Since ; a = O, mod ^ t h i s condition i s equivalent to 5 =_ mod 4* . By the hypothesis of Case I I there i s no maximal reduced integer of the 1st degree i n and therefore the conditions a = i . , mod u- and 5 = mod , are not s a t i s f i e d . Hence we have the integer ' 0 > x) provided S == Ql—I, mod # , and provided the conditions are not s a t i s f i e d . I f ' °> ~k ) i s not maximal there must be an integer of the form ( > ~p ' J where t <c Howevers before we can discuss t h i s general integer, i t i s necessary to f i n d the conditions for the existence of the integer J =. (j^ ' 0 > J^) Since the existence of £ implies that of (j_ > & > f ) we have £ = (Q.-/),mod H and we may therefore write S ~ Q.-J -f-u-4 Also since 770 = ji w e bave & 0 = -^ 7 We f i r s t dispose of the case £ e = -/ . From _Tj ^ , on w r i t i n g 5 = Gl.-I we get Hence, i f Cl = 7-Q. Q_ 4- I = 0 ^i^ati if Therefore Ql i s odd, and Cl ~ i. u - • (3) ana . / c — ' . Subst i t u t i n g i n 7~z and- 7", ; the former i s s a t i s f i e d , while the l a t t e r gives I -f- A +. 4.-S S = Oyyy^Ci i-x & Or (/ * A+S) Qstousd % whence 5 ft) But (3) and. (4) are the conditions f o r the existence of a maximal reduced integer of the 1st degree i n and hence, by the hypothesis of Case I I , they cannot be s a t i s f i e d . The integer Jf t therefore., cannot e x i s t i f CL0 = - / _Tow if. do — I we have from Is Therefore (Ol-x) 0, mod % } and since i t i s a perfect square, (_»-2_)'L = Qt mod /6 Eence we have 6=_ ,2,mod ti and We may now write S = GL- / $ a n d su b s t i t u t i n g we f i n d that J. _ i s s a t i s f i e d , while X gives / - l f l ^ c i ^ ' i s - j f t s -f-Q% ~o?w*d 1 or >••. v 4, . _ % (/ ~ a + S) = O^p^cii^ Z whence 5 f6 We therefore have the integer (jz > & > J*-J the conditions f o r I t s existence being Q — i . , mod //. } and S = 6a mod /6f We now consider the general integer where t < ^ F i r s t we take the case ^ 0. From we have '^W^L Hence i f ^ -1 1 , £ cannot be z / greater than since then X ^ would give r i s e to - ^ j * the integer fJ- , r> -L\ f o r which we must have _ cannot Q. ^ X r mod. H- and. therefore since X I£t t be greater than ^ 2-. f-.sM. I f ^-z* =/ } which i s the only other p o s s i b i l i t y , t-r ' • • •• 1 .^ gives r i s e to an integer j , ~ (~r •> x > , We s h a l l now prove that t h i s cannot e x i s t . The existence of implies that of (j. } o} ~) and hence we have 5 = -(6LH)mo&. tt and we may write £ = uiz ~Q.-/ . Also since - £ < ^° __ •£ we have a<, — */. I f o-o — I. . substituting i n JT3 we get and therefore mod and We may therefore write -S = % - Ql-I , and substituting we f i n d that Tx i s s a t i s f i e d , while J, gives •••/ * + -f-H-S-f- 2. Q.S +S'<-.•—. Oy^uw( X . whence 5 _= This i s impossible since, by hypothesis, there i s no maximal' reduced integer of the f i r s t degree i n ~xr. Now i f a.0-=.-i} from I3 we get or i f « = i « / Cl* +• i + x-A == o^pt^d \ lt yk^Q c^-fJU^ CMM iAe^ n-t~1 Therefore &' i s odd and Q = / J mod ^ Henoe Q.s 2. mod h , and ^ i s odd. Substituting i n I*. and I, the former i s s a t i s f i e d , while the l a t t e r gives Substituting 5 = U-k and Q = i f t ' we get - ^ 4 — ^ b A ^ - i s - i ^ ^ e e i b i e - j — f o r i — s i n c e — ^ — a n d : ^—-are—be^tb CP^£^ e-dd-i—every- term—esc-gpi;— r s — d i v i s i b l e by— y J t ^ ^ ^ ~ Irenes there can bo no ini-^ge-r > ; 1 J^)- __ "V^" Tho-?ofore v;c cannot have ^-=-f- and therefore ^-t Si L A\ -ahgr?-&®-~^teow& abOV-e^ - The Integer ^ therefore becomes • ' ' /a / ^ I f ^ i s an integer then t f. = (-ji > 0 > If) i s also an integer. Therefore, since an integer > o , -AJ e x i s t s , but 7 0 > 'j}) does not, we must have G0-~ J mod . Substituting ^ i n -Z^  we get // Co ~ i. A = O svwiHX X '•n-f and therefore ia0 = a + X""a . On substituting t h i s i n I3 we f i n d Therefore £ 1(0.*-us) and we may write Q?~-ifS = ZT'^ and, sub s t i t u t i n g t h i s we get T h e r e f o r e s i n c e Q = jLj mod 4 we have (s) S i m i l a r l y , s u b s t i t u t i n g i n 1^ and w r i t i n g XO-c =.&+£. a n d Gl-hs ~t we g e t a f a ' ^ a - ^ j .^ 2 o (W/^ (6J and a g a i n , from J . , ( A x * -4-) + 8-£a - + X a-ls * -"""(7) F r o m (j>)» i f a. i s o d d , •£&•-•/ mod 4 and hence t l - ^ = Z ; mod and t h e r e f o r e T h i s i s i m p o s s i b l e , f o r from (7) s i n c e A ^ 5 T h e r e f o r e a must he e v e n , and from ( 3 ) , mod ^ H e n G e 0 " - ^ 5 = O ^ W l 1 " and l&o ~z&,/yvurii £. T h e r e f o r e , s i n c e ^ i f we l e t a = X-O! we have £•<> S Q. mod # , But i t was shown above t h a t a 0 = /,mcd # and t h e r e f o r e , - 3 1 -a ' _ f -Since _= 3 congruence (7) now becomes or i f we l e t CL = £ tf/, a ~ £ a ^ a . n d ~ u4-' (V W - ^ ' J S 0;swd/(, - #J low It -4- i s even 0^ mod if and there-fore, since i t i s a perfect square and therefore a'x + &' S ^ ^ / ^ • " -But t h i s i s impossible since a' == mod A4 . Therefore ' cannot be even. From (9), since i s odd + a'-•£').&.. 0j-wrf-i. Since Q and are both odd, a must be even and hence ft--£'^0 mod^ and since, from ( 8 ) , a ' == / ; mod ^ w e h a v e and we may write ' = tf-l — l and since #/ i s even, Hence i f C - ^ S = . 4 . T V ^ ^'^and i f Q~ i mod. $ then there e x i s t s a maximal reduced integer •••) j •» / where J_a6 = fi mod 1 ^ the only ease l e f t to consider i s when at - o and ±. that i s -when there i s an integer of the form . f = (% ' ° ' i^.y ^ > J L As before, since Z ^ = ° ? ^J we must have a6 ~ / mod l+. ^ we have £ a 0 = Q., mod £ . • and therefore ;t£ 6 =• Cz *• i.^ <X , Substituting t h i s i n 1 3 we § St /-n-i i i X-n + l 3-X. O. - + US S . Ojsnwoi t i.K--!L ly % . ^ 3.-*.-% /) Therefore 2- I(& '^SJ a n x i w e may write Q.-US — J-whence the above congruence becomes 3 ^ -4- ^ Oy^>^§' (/O) Substituting s i m i l a r l y i n J we get the condition and from J, ; 4 - a .-. = -.o^yyv^d/6 -- " 0 0 From (10) we have ^ s= ^ + J CL% and sub s t i t u t i n g i n (11) we get / ^ Is It (v ••• 2* y (7J 1 J^L^^-e^. I^ju-ijl- JL-ya-A-'£r * i^^Z—tj^^ y) - 33 -Hence a. i s even and we may write a = 2. a\ whence Therefore a.' and are either both odd or both even. I f G-' and A are both even, U I cl and from ( l l ) , /6 Hence xn t h i s case Q_ ~fys ^= ot mod 4. • and i £ 0 = a m o i i . I f a' and are both odd, we have from ( l l ) , #/-t£ but and we can therefore write 4-- *f4 and therefore since •.•=•./_, mod ^ we have = ••mod 44 , m this case, therefore, GL -4 S = X \tf-<t+0 and 4<20 == &, mod £ „ Z £ 0 ^ ^mod In both cases, since S we have 5 ^[>mod ^ and therefore, since = / mod # , Q. = 2 , mod Hence i f Q =. mod g', and i f a = ^0*^*0, there i s a maximal reduced integer ^~Sv J 0 .> J"-*1} where i.ao =. Q., mod x. , but ^ <3_, mod j i . I f fl - ^ 5 . i s not of t h i s form, there i s an integer (~~^ > 0 > j f ^ J where ^ i s the greatest integer such that X /09- " However, i f GL~-h$~ 1*''** L(^u<4.-f)} the l a t t e r integer i s not maximal since i t has been shown that i n this case there /Go j_ j_ \ i s an integer / ~j~+t } %_ } ] • This completes. Case I I . The following table summarizes the r e s u l t s , that i s , i t gives a l l the maximal reduced integers of the 2nd degree which can occur when there i s none of lower degree, together with the conditions necessary f o r t h e i r existence. - 34 -TABLE I. Maximal Reduced. Integers # rL. Sec^Uh^ywsL"" Conditions f o r Existence Sub-case A; £lQ. and//s 1 — / / a . a n d / V - S , ( 2 ) , ( 3 ) , and ( 4 ) ; ; u n s a t i s f i e d . 2. i i . V X ^ s =- O,mod > ^1. >: 4. 4. ^ + 1 Sub-case B; ^ i s the greatest integer for which ^i"^0 * t h i s congruence holds and 4a 0 ^= a.mod^"", ^ ^ . & and S s mod. /6 , #9 u ^ ^ ^ ^ > l . i-/ft andz Y s , a = />smod£ , ^  > ± - I f a*-*sis of the f o r m^ ( W - $ > , ^  = and^.sa mod X", but"^ Q, modX^*'. But i f e x - « i s not of t h i s form, then -m. i s the greatest integer such that,•.•£*'"**•/<&*'-and = Q, mod i ^ f the greatest integer f o r which t h i s congruence holds and za„= <2, mod ft™ Sub-case C:  ^ / S but ^  There can be no maximal reduced integer of the 2nd degree i n t h i s case. 7 ' - 33 -Maximal Reduced Integers ^~pL5^^ J>^y^-- Conditions for Existence Sub-case D t j/q but 36^ ,5 G=(s*/),mod^ ( 7), ( 8 k (-9) and (lo) u n s a t i s f i e d , and the conditions mod*, and s = mod/6 not both s a t i s f i e d . • : • /.•• •• •. / " " ; " ' : & == £ > m o d a n d s == (a-/j mod/6, (8), (9) and (10) unsatisfied-8« — +• i -x +-h, Q and = l} mod & , q^-hs - x'(n-t-i) , svx > i } ZG-o = C, mod z' Q=l,mod 8-. , I f G.x~»S=l (ttJ+t), sn>i, and zo.0=q mod A."1, but ^ Q mod I f ft— ks = + , •>*.. i s the greatest integer such that i.7""**-1(0?- »$) and o.fl0 == a mod %T*! ( a^ to) G = i,mod 8, a - # 5 = i _ ~ * ' Y ^ * ~ ) and j. a 0 _= 6,mod £ . y 2*- Z. £/• J - 36 -3• Case I I I , We s h a l l now f i n d the maximal-reduced integers of the 3 r d degree i n when there exist none of lower degree i n ^ • The work i s divided into the same four sub-cases as Case I I . In addition to Theorem ¥ and i t s C o r o l l a r i e s , we require the following: Theorem B» I f a maximal reduced integer i n -f of the 3 r d degree i n exis t s , but none of lower degree i n then there exists a single-prime reduced integer such that a l - a , = -Q j • Since there exists a maximal reduced integer i n $ of the 3 r d degree i n -as then there exists a single-prime reduced integer of the form Reducing ^ by the f i e l d equation, we have f t P f ^ Subtracting t h i s from a * £ we get o o f f j> Since t h i s is an integer of the 2nd degree i n -x i t must reduce to ordinary integers, and hence ( G-, Cl^ ~ ao f sTrurd. f as the theorem states. Sub-case A; d> /Q- and $/S e By Theorem V, Cor. 1., i f there i s a maximal reduced, integer of the 3 r d degree i n ^  but none of lower degree, then there i s an integer On substit u t i n g , we f i n d that and J-ff. are s a t i s f i e d , while I, requires the condition that ft^ I £ However, from Table I, # i f ^ and ^ V s there i s an integer - of the 2nd degree i n -*c which i s contrary to hypothesis. Hence, i n t h i s case, there can be no maximal reduced integer of the 3 r d degree i n -x . Sub-case B: •£> j>& and j> 4>S. In t h i s case, by Theorem Y, Cor. 3.„ i f there i s a maximal reduced integer of the 3 r d degree i n but none of lov/er degree, there is a single-prime reduced integer where <X0^ o and O. Also, since from the second con-gruence of Theorem B, a, ax = <a0,mod ^  we must have CLt =£ O . 38 i f f ^ , r,v which i s impossible since/ G ana S are/b^oth odd* Hence we musjr have /> 7^  i. • Substituting ^ i n and multiplying the r e s u l t i n g congruence by Q we get cx, o.x % as - a3 = 0,/yyi&vl f> and from 60, u ( s - a, a * i"-&5 - 3 Q s = D, adding ; j.cl,S - Q.S == 0, ^ or, since ^ . Id, ~ Q.}yO^(f> A l s o 9 from I\ } since .2 &o = a^Q.yrr^^ We may, there fore 9 write zcz0 = CL^Q. and ±a, — Q. Also we have / /6 S (a*- ^s)*" and since and ^ jfe /(Gl*~-sj Ho higher pov/er of ^ than the f i r s t can divide Qx-f*$ ^ f o r i f Q.*"- £S s 0_> mod ^ * then from Table I, there i s an integer ^—- o > ~ J of the 2nd. degree i n . We, therefore, have Q.2"— hs , where -c i s prime to ^ -How, s u b s t i t u t i n g f o r S-&0 and z A , i n X ? we get - xa^(ax-us) + &(aL~ tts) - X j> 4 * s) = 0,^yoL-f^ or, since a*"- US .= #•<. and f> 4>-<-%. CLi — d = o , sm<a?i jf) 0) S i m i l a r l y , from I^ we get &x C ^ s > -i- a a. x(g£-~us) -1<x~^c(c£~~ h9) which., on sub s t i t u t i o n of Q. - ^5 = reduces to whence, i n view of ( l ) , or since a x -i_aii = O, ^^cLp Making the same substitutions i n J , we get which, on writing Q.^- i+S = ^-c . ' becomes whence if % - 40 -Nov?, from (1), la? =• Q., mod ^ and ^ a ^ s mod |5 Substituting these we get But t h i s i s impossible s since a -k$= O mod ^ , and adding these two we get * f l l s O,, mod ^ which cannot hold since $ ^ %• and $ 4> a • Hence there i s no integer ~r (Ci0 , d, , a.^ , f) and therefore there can be no maximal reduced integer of the 3rd degree i n -Sub-ease 0; 41$ but & ? Q-I f there i s a maximal reduced integer of the 3rd degree i n -x but none of lower degree, by Theorem B, there i s a single-prime reduced integer ^ = ~p (&o > o., j ax i i') such that a0 ax" = - S j (0 a, az = a0 V/rwcLf (ZJ a ? - * , * - * J ( ? ) From Theorem V , Cor. 2 . , £ 0 = 0 , and hence, from (2) either C L i - q or a,^ - o but from (3)» since jfr4>Q.} CL, and cannot both be zero. The only possible integers, therefore, (cL) >h-tnd Ifj S = f c O Tn^d • (hj f ^ ' V -n^-o-d j S B -TH-ereS If} 6 . ^ . - / -^^Y-y ^ l<^j£c g = •*-&,•<-/ ~ ^ ^ 2 . . ^ / , ^ ^ ^r-c^^-j S \c 0 rt-^rrl 2. • 5 £ l>- y^r*{• If V - 41 -are JL<>, o, a.x , / ) a p p and. f (o } a, } O, i) , a * ». The f i r s t of these, however, cannot occur. For substituting Pa^-i n we get 2>(i. s - Q / = C?, mod. ^  which^is impossible r since ft IS but ft 4>(X-We need., therefore , A only consider ^ - (o, a,/J Substituting i n we get = Q^ mcd ^ >, or since ^/S and ftf-CL a, = a mod ^ ; and therefore a, = a +• -kf-Substituting t h i s i n I3 we f i n d that £ = Oj mod ^ * and X, , _Ta , and 2"^  are a l l s a t i s f i e d by t h i s same condition. Hence there i s an integer , G-, j O, /) i f and o-,= Ql mod -ft>. I f t h i s i s not maximal, there w i l l be a maximal reduced integer of the form , where ^ >s3 and sm We consider f i r s t the case fi^fi±. From Theorem I we have ft***I/6S(Q.x-fys)X and since ^ / f i but ^ ; ^ j>(Q'L~/ss)J&ntiL hence for -p ^  z , i t follows that ^ * 7 £ . Also from ZS(P~) since S and ^>^Q. •we 'have £, s Q._, mod ft*, and therefore — GL + p^Ji We have or, since ^/£ and ^^/(a,-a) - 42 -apart from ordinary integers. Therefore <o = -^.jfor i f not •p ^ -7c where i s the greater of and ^x. gives an integer ^ o r 'T^--* ^ > according as > ^ or ^ < -a,and as. neither of these can e x i s t we must have <d = -t. Hence Substituting i n I¥ } we get X d0 - a = £>, ^vo^L p - ^) and from Is(f) -3a0Q+ata\taxS= Oj sryu*>( Since ^ and therefore, since + axa = -A adding t h i s to ( 4 ), we get and therefore a0 ~ o and from (4) a % = o. Hence the integer reduces to Substituting t h i s , J. and 1", are s a t i s f i e d . •J" gives 5' -Q (£- -*/) V- 5 ( W , -3Q.) = O)sry«rd'£ which i s s a t i s f i e d since £ f S and a,- &~ 0; mod ^ T S i m i l a r l y gives which i s also s a t i s f i e d . Hence i f ft-f ±, we have the maximal reduced integer - 43 -( ® } y ® J where sn. i s the greatest integer such that ftx"IS and. a, = Q_ mod ft™ We now consider^the case ft ^ £. • I f {0}-^yOj^J ^ i s not maximal, there i s an integer of the form °^^<*i& wher^ ^i^"^3 and . From Is(£) i f ^  we have or - 3 Q a o -f £* \^ <2 V *. S) £ 4 •= Oysrhfret 1*''-—-C6) according aA y^ or ^ Neither (3) nor\(6),ean hold unless f o r i n each, one term is d i v i s i b l e by while the other \s not. Hence we mdst have e i t h e r ^ = A or a0 = * f c = o . \ r^^yJUrdJ F i r s t we consider the $4?om J.s(dt-) ^ -jft -3Q cl0 +C&%+%&)&z=0,'^*^%.'% and e i t h e r aB ^ o and a^^. o or a0 — a*_ = o . ±s-^phe l a t t e r case i s considered l a t e r 0 ^ we—44rS*ega*d—i-t~he-3?-e-. Substituting i n .!> we get Since a0^o and a%^o t h i s i s impossible unless — / Hence the only possible integer i s from which But (_0 , x ) O j -%_y i s also an integer, and subtracting th i s from ^ we get an integer (- s O > of the 2nd degree i n ^ which must reduce to ordinary integers. Hence we have i- l=> and c,- Q. = 0, mod 1 -Substituting ¥ i n i 3 we get a(&,-Q.) zi~~*'Q.i'+ us (a,-a) + dS = O, ^nurd X , and therefore, since XJS and £^l(at-G) and X $ O. 4. « * S = <?, smovL X . Hence i . /£ and i f 5 = 1 -£ , we have £E — Q. ^ — $0' S i m i l a r l y from I * we f i n d Z- 3 & + a x + Q$-'Q^s'^ o and since // mod 4-^  -asd~*»©ffl-44)——^-W^^se-d-^, we—get— eaad- J J L i s evidently s a t i s f i e d . Substituting i n I, we get 1 (&-/)• • • JL Q(a,-Q) + X S(a,-Q) Writing ft, = Q + z and 5 = i - t h i s becomes Z. (l + QSj -hi Q.4L-+ X A Q S f l S[X &A +x-£•] -hi as ==OymuytiX - 45 -In view of ( 7 ) , we may write s' = */-Q and therefore / f f is ' = and since Q 1 = /mod >^  we have I + Q.9' = O, /rrurct H-Hence the f i r s t term of (8) i s d i v i s i b l e "by and may be omitted, and the congruence becomes i 1 A Q. + 1 4Q £ •/- Z. 5 {a,4 + L s' j + X QS = 0,/moxLl We now substitute 5 ' = M-€ - Q. , and on d i v i d i n g by • -2. we get or since a.,x -J^^uM^f, f-i d) =0}^v<rc(^ which i s evidently s a t i s f i e d . HenceA i f S X. 5 where S s mod there i s a maximal reduced integer ^ > - j - , ~ > j-^J where We now consider the case £<, = <3.x = o If)there *<-~^  is an. Integer ^ then . . / 1 ar r #••<,.,..•:.: i s of the 2nd degree i n and hence must reduce to ordinary integers. Therefore we have ± JS and a, = Q, mod £., Substituting i n 13 , w e have Q. (&,~ a)x + us (a,-a) + as ~ o,srrt^ il*' and therefore,, since 1~JS and a, = e.moa and i t follows that Similarly,we f i n d that 1, i s s a t i s f i e d by the same condition while Jj_ and 2% are i d e n t i c a l l y zero. Hence we have the integer (o } ~^ > 0} j^} where --n i s the greatest integer such that iV""/£^ and a, = & modi.'"- This w i l l only be maximal when the integer f£ > -ri •> > -p,J f a i l s to exist,^ since i n the l a t t e r i s the greatest integer SUCh that - A-i^o , i-^^iJLla. CA^L, l^-e -t^i^^ AJZ^-Z_ (2-^/ /JJy^. ^^Cu^Cl^^ ^a 1 ) ^Pfrj^-e. ryu^i^aj? /<z^l~c*f»^4 ^^C^iZ^s Lst_ Sub-case D; j/Q but^ ? 5 . From Theorem B, i f there i s a maximal reduced integer of the 3rd degree i n but none of lower degree, there i s a single prime reduced integer / = f ( a ° > a> 1 a± > 0 such that &o <XX = - S snwrfft> Q) = do ^nvtrd. -ft - (Z) t and since fa &x - a, = o ^y^ol^ 0) From Theorem 7, Cor. 3., ( or from congruence ( l ) , above), we have a „ ^ o and G.x^t 0} and, therefore, from either ( 2 ) or ( 3 ) , we have also &, =£- O From Theorem I, we have and therefore since ft I OL but , we must have jf)= z • - 47 -Hence a0 — &, = a.<x = / and the integer becomes 7 * (i • i J i ' i) Substituting ^ i n I,} ZH we f i n d that I-f i s s a t i s f i e d , while gives z -f-Q + 45 -fas ~=- 0, ^rt^trt/ /f OX!* (%. -f- a) (i + s) = o} ^vwcrfC h , which i s s a t i s f i e d . From 2 4 , after c a n c e l l i n g and omitting terms d i v i s i b l e by 8, we get **(S*+r) s 0,mod S" which i s also satisfied,. Substituting i n I, we get or, since X IQ. 5 • and therefore / * S f d £ £>, s**uro( tf whence, since 1 M , S IB. Ct-Zyn^fffff-However, from Table I, t h i s i s the condition for a maximal reduced integer j * f rtX of the 2nd degree in ^y^x^w and i s therefore contrary to the hypothesis of Case I I I . ^ / - ^ Hence i f but jfi $S there can be no maximal reduced integer of the 3rd degree i f there i s none of the 2nd degree. The r e s u l t s of Case III are summarized i n Table I I . TABLE I I . Maximal Reduced. : Integer Conditions f o r Existence. Sub-case A: it>lQ. and//£. There can he no maximal reduced integer of the 3 r d degree i n x i f there i s none of the 2nd degree. Sub-case B: & £ a and £ j>£ There can be no maximal reduced integer of the 3 r d degree i n -x. i f there i s none of the 2nd degree. Sub-case G: I Is but i j>Q r r . •• f^Xj /yt i s the grestest integer such that ^Z^/s, and a, = 0,mod^^ 2» • . 3 S — i . ^ , -<n = X, where 5 S - a . 1 1 x mod t+, and = Q_ mod X . • I. aA0}&-5/ (2) unsatisfied. ^ the greatest integer such that l ^ / s and modi."! • j Sub-case B: but ^ 5 " . There can "be no maximal reduced integer of the 3 r d degree i n i f there i s none of the 2nd degree. 

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