On a Fermat-Type Diophantine Equation

Transcription

1 Journal of Number Theory 80, (2000) Artcle ID jnth , avalable onlne at htt: on On a Fermat-Tye Dohantne Equaton Sankar Staraman 1 Deartment of Mathematcs, Howard Unversty, Washngton, DC Communcated by A. Granvlle Receved June 30, 1998 Let >3 be an odd rme and a th root of unty. Let c be an nteger dvsble only by rmes of the form k&1, (k, )=1. Let C () be the egensace of the deal class grou of Q() corresondng to, beng the Techmuller character. Let B 2 denote the 2th Bernoull number. In ths artcle we aly the methods (followng H. S. Vandver (1934, Bull. Amer. Math. Soc., )) whch were used by the author (1994, Ph.D. Thess, Calforna Insttute of Technology) to rove a secal case Fermat's Last theorem, to study the equaton x + y = c z. In artcular, we rove the followng: Assume s rregular, and B &3.Letq be an odd rme such that q#1 (mod), and there s a rme deal Q over q n Q() whose deal class generates C (3), whch s known to be cyclc. If x + y = c z has nontrval nteger solutons, then we show that q % ( c z (x+ y)). We also gve roof of the unsolvablty of the above equaton for regular rmes ( >3), usng the results of H. S. Vandver (1936, Monatsh. Math. Phys. 43, ) Academc Press INTRODUCTION Let be an odd rme. Snce the roof of Fermat's Last Theorem [W, TW],.e., that x + y =z has no nontrval solutons, t has been known that [R, DM] the same technques can be used to rove the unsolvablty of equatons of the form Ax +By =Cz, where A, B, C are constants. But the same s not true for equatons such as x + y = c z, when there are a large number of unrestrcted rme dvsors n the coeffcent c. In ths aer we consder the above equaton when >3 and c s dvsble only by rmes of the form k&1, (k, )=1. Let a th root of unty. Let C be the -Sylow subgrou of the deal class grou of Q(), wth order h.letc + and h + denote the -Sylow subgrou of Q(+&1 ), and ts order, resectvely. The Techmu ller character : Z Z * s gven by (x)#x (mod ), where (x) s a 1 I thank D. Ramakrshnan for suggestng ths roblem and hs encouragement. I also thank L. Washngton for hs valuable comments, and L. Sharo and F. Ramaroson for helful conversatons. I am also grateful to the referee for hs valuable suggestons for mrovng the aer X Coyrght 2000 by Academc Press All rghts of reroducton n any form reserved. 174

2 ON A FERMAT-TYPE DIOPHANTINE EQUATION 175 ( &1)st root of unty, and x # Z. Under the acton of ths character, C decomoses as a drect sum of C () (), where C s the egensace corresondng to. It s known, by a result of Kurhara [Ku] (also ndeendently observed by Greenberg), that C (3) s cyclc. The man results of ths aer are the followng. Theorem 1.2. Let >3 be a regular rme, and c an nteger dvsble only by rmes of the form k&1, (k, )=1. Then x + y = c z has no nontrval nteger solutons. Ths theorem follows easly from the results of [V4]. It can be shown that the theorem s true for certan rregular rmes as well usng the same methods (cf. [Wa, Cha. 9]). In artcular, for such that % h +, 3 % B for all even &3. The followng theorem also concerns rregular rmes, but wthout the assumton that % h +. Note that, n any deal class of Q(), one has an nfnte number of rme deals of degree 1 by the Tchebotarev densty theorem (cf. [C,. 257, Wa,. 333]). So gven an deal class that generates C (3) there are an nfnte number of rme deals of Q() n ths deal class whch le over ratonal rmes congruent to 1 modulo. Theorem 2.1. Assume s rregular, and B &3, and c s as n Theorem 1.2. Let q be an odd rme such that q#1 (mod), and there s a rme deal Q over q n Q() whose deal class generates C (3). If x + y = c z has nontrval solutons x, y, z # Z, then q % ( c z (x+ y)). Remark. As mentoned n [BCEM], there are only two rmes below 4 mllon, namely and , for whch B &3. But the roof of the above theorem can be extended to B &5, B &7,... and so on to many more ndces (exactly how many deends on ), assumng that for each ndex & the corresondng C () and C ( &) are cyclc and trval, resectvely. Our roof of ths result follows, wth many modfcatons, the methods of [S]. In [S] we used a seres of deas of H. S. Vandver [V1, V2] along wth a theorem of M. Kurhara [Ku] and some consequences of the roof of Iwasawa's man conjecture for cyclotomc felds due to B. Mazur and A. Wles [MW] to rove a secal case of the FermatWles theorem. After recallng some notatons and relmnares, we rove some elementary roostons about x + y = c z whch effectvely render the roof of ts unsolvablty amenable to the same methods used (classcally) n the second case of Fermat's last theorem. Indeed, the roof follows easly from the results of Vandver [V4] and t s ncluded for the sake of comleteness.

3 176 SANKAR SITARAMAN The roof of the man theorem then uses a modfcaton of the methods used by Vandver [V1], as descrbed n [S2, Cha. 3]. To comlete the roof we need to show that certan generalsed VanderMonde matrces are non-sngular mod. ths s done by establshng a congruence relaton between these determnants and the Bernoull numbers. 0. NOTATIONS AND PRELIMINARIES >3 s a rme, s a rmtve th root of unty, *=1&, and ;=(1&)(1&&1 ). K=Q() s the cyclotomc feld, A=Z[] ts rng of ntegers, and A* the grou of unts n A. G=Gal(KQ), _ s a generator of G and _()=r, where r s a rmtve generator of (ZZ)*. C s the deal class grou of K, h the order of C. C, h s the -Sylow subgrou of C and ts order, resectvely. K + =Q(+&1 )=Maxmal real subfeld of K. A +, C +, h +, and C + are defned smlarly. If I s an deal n A, then [I] s ts deal class n C. The Bernoull numbers B m are defned by t(e t &1)= B m=0 m(t m m!). The generalzed Bernoull numbers B m, j are defned by &1 a=1 ( j (a) xe x (e ( &1) x &1))= B m=0 m, j(xm m!). The Kummer Unt $ s gven by $=- (1&r)(1&&r)(1&)(1&&1 ). The cyclotomc unt E m s defned as E m => &1 k=1 (_&k ($)) rmk. Note. Ths defnton of E m s not standard. For nstance, the E m defned n [R,. 128] s actually the square root of the E m defned above. But all the roertes of E m that we need wll follow easly from those of - E m and vce versa. For a rme deal L{(*) ofa, and : # A, : L, theth ower resdue symbol [:L] s defned by [:L]=a, where a s the unque th root of unty such that : (N(L)&1) #a (mod L), N(L) beng the (absolute) norm of L. [:L] s extended to all deals rme to (*) multlcatvely. Also, we have { : J= \K(:1 )K :1 = J + (:1 ), where J s any deal of A wth (*) % J, and ((K(: 1 )K)J) s the generalzed Frobenus element. 1. PRELIMINARY PROPOSITIONS Throughout ths aer, >3 and c # Z s such that only rmes of the form k&1, (k, )=1 can dvde t. In ths secton we rove some

4 ON A FERMAT-TYPE DIOPHANTINE EQUATION 177 rearatory results about the solutons to the equaton x + y = c z.we could also assume, wthout loss of generalty, that c s not dvsble by the th ower of any nteger. Prooston 1.0. If c s as above, and a rme l dvdes c, then [(l)]{1. Proof. [(l)]=1o 2 (N(l)&1)O 2 (l &1 &1) whch s mossble when l+1 s dvsble by but not 2 as n the case of rmes dvdng c. Prooston 1.1. If X +Y =&; m cz s satsfed wth X, Y, Z # K +, where c s as above, (1&) and (Z) are relatvely rme, & s a real unt, and m(&1)2 then (a) (X, Y)=1 (b) The factors X+Y, 0 &1 of X +Y are all arwse relatvely rme, excet for (1&) whch dvdes all of them. Also, (X+Y, (X +Y )(X+Y))=(1). (c) Let l c, l a rme number. Then l % ((X +Y )(X+Y)). Proof. (a) If (1&) dvdes (X) and (Y) then we can dvde t out of (; m )=(1&) 2m. Suose L{(1&) s a rme deal that dvdes both (X) and (Y). Then L (cz ), whch means (because we assumed c s not dvsble by any th ower) L (Z), contrary to the assumton that X, Y, Z are relatvely rme. The roof of (b) follows easly usng (a) and the fact that % (X +Y )(X+Y) (cf. [Wa,. 168]). (c) Now, let l be a rme dvdng c, of the form k&1 and let L be a rme deal of A that dvdes l. Then l (X +Y )(X+Y) O L (X+Y) for some {0. Snce l#&1 (mod ), the decomoston grou of L s of order two, whch means t s generated by the conjugaton automorhsm. So L s fxed by conjugaton and hence L (X+&1 Y) whch combned wth L (X+Y) gves L (X) and L (Y), a contradcton. Suose x, y, z are relatvely rme ntegers such that x + y = c z. (1.1) Lettng Z=z t (t=v (z)), &= t+1 ; m, m=(t+1)(&1)2 ( &1)2, we can aly Prooston 1.1 to get, (x, y)=1 and wth u, v, t # Z, (u, v)=1, (v, )=1, (u, )=1, x+ y=c t u (1.1a) x + y (x+ y) =v. (1.1b)

5 178 SANKAR SITARAMAN Equaton (1.1b) can be wrtten as &1 =0 \ x+ y 1& + =v. (1.1c) The deals ((x+y)(1&)) n (1.1c) are rme to each other and (*). Thus we have \ x+y 1& + =I, where I an ntegral deal. (1.1d) Remark. Ths reduces the roblem amenable to the classcal methods used n the study of the second case of Fermat's last theorem. In fact, usng the results of Vandver [V4], the equaton x + y = c z can be shown to have no solutons when s regular and >3. The roof s ncluded below for the sake of comleteness, and t s along the lnes of [Wa, Cha. 9]. Moreover, usng those methods we can rove that x + y = c z has no nontrval ntegral solutons f % h + and satsfes assumton II of [Wa, Cha. 9]. In artcular, for such that % h +, 3 % B \ even &3. Theorem 1.2. If >3 s a regular rme, c as defned above, then has no nontrval ntegral solutons. x + y = c z (1.2) Proof. Assume there s a soluton. We wll get a contradcton usng a descent argument as n [V4, Wa, Cha. 9]. Let Z # A + =Z[+&1 ]be such that t has the mnmal number of dstnct rme factors among all solutons of &1 =0 (X+Y)='B; m Z, (1.2b) where X, Y # A +, ' s a real unt, m ( &1)2, and the only rmes dvdng the nteger B are of the form k&1, (k, )=1. If (1.2) has a nontrval soluton, then (1.2b) has a nontrval soluton, by lettng X=x, Y= y, Z=z t (t=v (z)), B=c, '= t+1 ; m, m= (t+1)(&1)2( &1)2 where t s as n (1.1a). From (1.2b), t follows from Prooston 1.1 that for =1,..., &1, \ X+Y 1& + =I, where I an deal of A. (1.2c)

6 ON A FERMAT-TYPE DIOPHANTINE EQUATION 179 and (X+Y)=B(;) t( &1)2 I 0, where I 0 s also an deal of A. (1.2d) The deals I are rme to each other and to (1&) for0(&1). Also we have (Z)=I 0 I 1 }}}I &1. These deals must all be rncal because s a regular rme, and from (1.2d) I 0 n artcular can be seen to be of the form (\ 0 ), \ 0 # A +. So we have X+Y=' 0 ; t( &1)2 B\ 0. (1.2e) Here ' 0 s a real unt and \ 0 s rme to ;. Now, n [Wa, ] t s shown that, gven cyclotomc ntegers X, Y satsfyng (1.2c) and X+Y#0 mod(1&), we have \ X+Y 1& + =' \ for 1 &1, (1.2f ) where the ' are real unts and the \ # A. Note. Under the condtons satsfed for X, Y, Z here, we can also get (1.2f) usng Prooston 1.0. Now, changng to & n the above equaton (and assumng \ =\ &, then multlyng the resultng equaton wth t), we get (X+Y) (X+& Y)=X Y (+& ) XY=; ' 2 (\ \ ). We also have, from (1.2e), X Y 2XY=' 2 0 ;2t( &1)2 B 2 \ 2. 0 Here ; =(1&)(1&& )=2&&&. From these two equatons we get &XY=' 2 (\ \ ) &' 2 0 ;t( &1) B 2 \ 2 0 ; &1. Now f j0 (mod) and j\ (mod ) (ths s why we needed >3) we get another equaton &XY=' 2 j (\ j\ j ) &' 2 0 ;t( &1) B 2 \ 2 0 ;&1 j. From these two equatons we get ' 2 (\ \ ) &' 2 (\ j j\ j ) =' 2 0 ;t( &1) B 2 \ 2 0 (;&1 &; &1 j ). (1.2g) We can wrte ; &1 &; &1 j =$$;, where $$ s a real unt. Then from (1.2g) we get, wth $ a real unt, \ ' 2 ' j+ (\ \ ) +(&\ j \ j ) =$; (t( &1))&1 B 2 (\ 2) 0. (1.2h)

7 180 SANKAR SITARAMAN Now, by alyng (1.2f) for and j we get ' j # ' \\ j \ + mod (1&) &1 because ' =((X+Y)(1&)) \ & #(X+((X+Y)(1&))) \ & # X\ & (mod(1&) &1 ) and smlarly for ' j. Thus ' ' j s congruent to a ratonal nteger mod, and because s regular, ths forces ' ' j to be a th ower (cf. [Wa, Theorem 5.36]). Now let 2 X 1 = \' ' j+ \ \, Y 1 =&(\ j \ j ), and Z 1 =\ 2. 0 Then from (1.2h) we have X 1 +Y 1 =$;(t( &1))&1 B 2 Z 1. (1.2) In ths equaton X 1, Y 1, Z 1 # A +,(t( &1))&1( &1)2, $ s a real unt, and B 2 s dvsble only by rmes of the form k&1, (k, )=1. Moreover, X 1, Y 1, Z 1, and ; are arwse relatvely rme because \, \ j, \ 0, ; are arwse relatvely rme, as shown above (see remarks surroundng Eqs. (1.2d) and (1.2e)). Thus (1.2) s another equaton satsfyng the same condton as (1.2b) but n ths case t has (Z 1 )=(\ 0 ) 2 =I 2 0 whch wll have a smaller number of rme dvsors than Z unless I 1 =}}}=I &1 =(1) whch would force (X+Y)(1&) to be a unt for 1 &1. Ths can be shown to be a contradcton (cf. [Wa,. 173]). Hence Theorem THE MAIN THEOREM Theorem 2.1. Assume s rregular, B &3, and c as n Theorem 1.2. Let q be an odd rme such that q#1 (mod), and _Q over q n Q() whose deal class generates C (3). If x + y = c z has nontrval nteger solutons, then q % ( c z )(x+ y). Proof. Remark: As noted n the Introducton, Tchebotarev densty theorem gves nfntely many rmes q wth the above roertes. We frst need a rooston. Let L be a rme deal of Q[], such that L (x+y1&). Then by Prooston 1.1b, L % (x+ y). So f L les over the ratonal rme l, then l (x + y ) O x + y #0 (mod l) O (&xy) #1 (mod l). From ths we get l#1 (mod). Excet for the last aragrah, the roof of the followng

8 ON A FERMAT-TYPE DIOPHANTINE EQUATION 181 rooston, followng [V2,. 217] s almost dentcal to that of Prooston 3.1 of [S]. It s resented here for comleteness. Recall that when n # Z, n$ s an nteger such that nn$#1 (modl). Prooston 2.0. _: # A such that If Eq. (1.2) s satsfed wth x, y # Z, (x, y)=1, then ( &1)2 \ x+ n$ y + =:. 1&n$ (2.0a) Proof. As shown above, we have \ x+y 1& + =I, where I an ntegral deal, (2.0b) when x, y are as n the statement of ths rooston. Alyng a Stckelberger tye relaton (cf. [L,. 13, Fact. 3]) wth n 1 =n 2 =1, and usng the fact that all rme deals dvdng (x+y1&) are of degree 1 (see remark above), we fnd that the roduct > ( &1)2 _ n$ (I) s rncal Alyng ths to Eq. (2.0b), we get &12 \ x+n$ y 1&n$ + =';, where ' # A s a unt, and ; # A. (2.0c) Alyng _ &1 to (2.0c), we get &1 n=(+1)2\ x+ n$ y 1&n$ + =' (; ). (2.0d) Multlyng (2.0c) and (2.0d), and usng (1.1b), we fnd that the deal (;; )=(v). Hence ;; =Ev, where E # A s a unt. Takng the roduct of (2.0c) and (2.0d) agan, we fnd that '' E =1. (2.0e) But we know, by a basc result, that '= g=, (2.0f) where = # A + =Z[+&1 ] s a real unt and g # Z.

9 182 SANKAR SITARAMAN From (2.0e) and (2.0f) we get = 2 =E &. Snce s odd, we can fnd ntegers a, b such that 2a=1+b, so that = 2a === b =E &a. Hence ==(= &b E &a ), and '= g(= &b E &a ). Lettng :== &b E &a ;, we get ( &1)2 \ x+n$ y 1&n$ + = g:. (2.0g) Let l be a rme number that dvdes c. Then by Prooston 1.1, l (x+ y), but l % (x + y )(x+ y). From ths, we get (x + y )(x+ y)=v # y &1 (mod l) O y#v (mod l), for some v 1 1 # Z. Now alyng the th ower resdue character [( )(l)] to both sdes of (2.0g), we get [g(l)]=1. If g0 (mod), [(l)]=1, whch s mossble for such l by Prooston 1.0. Hence Prooston 2.0. From (2.0a) we have ( &1)2 \ x+ n$ y + =:. 1&n$ Let q, Q be as n the statement of Theorem 2.1. Then we can assume, takng a conjugate f necessary, Q (x+y). (If the deal class of Q generates C (3) (3), then because C s an egensace corresondng to the egenvalue 3 (_) whch s rme to, the conjugate _(Q) wll also be n an deal class that generates C (3).) Then Q % (x+y) for{1. Now for k # [1, 2,..., ( &1)2], &kn$1 (mod), \1n(&1)2. Alyng _ & k to both sdes of (2.0a), we get ( &1)2 \ x+ &kn$ y + =: 1&&kn$ 1. (2.1) Now, (x+y)(1&)#(xe &1 e ) (modq), where we defne e = (1&)(1&), because Q (x+y). So from (2.1) we get ( &1)2 { Q=\ x e &kn$&1 Q +{ e &1 &kn$ = Q = 1 Q= {: =1. (2.2) Fx k 0 =&(&1)2, and let k 1 =&1, k 2 =&2,..., k ( &3)2 =&&32. Then we get ( &1)2 { e &k0n$&1 Q ={ e &1 &k 0n$ Q = ( &1)2 = { e &kn$&1 Q ={ e &k n$ &1, Q = =1,2,..., &3 2. (2.3)

10 ON A FERMAT-TYPE DIOPHANTINE EQUATION 183 Defne (mod ), nd(:)=a, where [:Q]=a. In ths notaton, we get from (2.3) that ( &1)2 : [nd(e &ko n$&1)&nd(e &k n$&1)&nd(e &ko n$) +nd(e &k n$)]#0 (mod ). (2.4) From [K,. 277], we have the followng relaton between nd(e ) and the nd(e j ), &32 nd(e )# : m=1 \\ 2m r &1+ nd(e 2m) 2m +&&1 2 nd(), where r s as defned n Secton 0. In (2.4), for any k, let &k n$= n, &k 0 n$=a n. Then ( &1)2 : [(nd(e an &1)&nd(e an ))&(nd(e n &1)&nd(e n ))] #0 (mod ). (2.5) Then usng the above relaton, we get ( &1)2 : _ ( &3)2 : _ (a n&1) m=1 #0 (mod ). 2m &a 2m n &(( n &1) 2m & 2m n ) r 2m &1 & nd(e 2m) & Smlfyng, and usng the fact that A t = def ( &1)2 (n$) t #0 (mod ) when t s even, we get &4 : u=1 u odd ( &3)2 \ : \ 2m u + nd(e 2m) r 2m &1 + A u(k u 0 &ku )#0 m=1 (mod ). (2.6) 2mu Lettng vary over 1, 2,..., ( &3)2, we get a system of ( &3)2 equatons n (k u 0 &ku ) whose matrx equaton has only the trval soluton f the followng generalzed ( &1)2_( &1)2 VanderMonde matrx A s nonsngular mod :

11 184 SANKAR SITARAMAN }}} }}} \ &1 A= }}} \ &1 3. b b b.. b 1 2 &4 3 &4 }}} \ &1 &4 Let the determnant of A=2. Then 2=\1 where. 1=} }}} }}} 2 & }}} 3 &4 1}.. b b b.. b b 1 \ &1 \ &1 3 \ &1 5 }}} \ &1 &4 Let B be the non-sngular (mod ) VanderMonde matrx gven by }}} B=_ }}} 2 &4 2 & }}} 3 &4 3 &2. b b b.. b b \ &1 \ &1 3 \ &1 5 }}} \ &1 &4 \ &1 &2 If det(b) sb, then 1=(x ( &1)2 )b where [x 1, x 2,..., x ( &1)2 ] s the soluton of the matrx equaton &. B[x 1, x 2,..., x ( &1)2 ] t =[1, 1, 1,..., 1] t (Cramer's rule). But B[x 1, x 2,..., x ( &1)2 ] t =[1, 1, 1,..., 1] t O ( &1)2 k=1 x k n 2k&1 =1, for, 2,..., ( &1)2 O ( &1)2 k=1 x k n 2k =n, for, 2,..., ( &1)2 O ( &1)2 k=1 x k ( ( &1)2 n 2k )= ( &1)2 n. Snce ( &1)2 n 2k #0 (mod ) for k<(&1)2, we get x ( &1)2 # ( +1)4 (mod). Thus 1 has to be nonzero mod and hence 2 s also nonzero mod.

12 ON A FERMAT-TYPE DIOPHANTINE EQUATION 185 Thus the system of Eq. (2.6) has only trval solutons. Consderng the leadng term only, we get ( &3) A &4 nd(e &3 ) #0 (mod ), (2.7) r &3 &1 where A &4 = ( &1)2 (n$) &4 as defned earler. We have the followng relaton between A j 's and the Bernoull numbers B (cf. [V3,. 114]): ( &1)2 : n &1 # 1&2 2 &1 B (mod ), 2, (&1) %. (2.8) Remark. The above congruence s also a consequence of Vorono's congruences (cf, for nstance, [R.. 108, 5B]). Let =4. From (2.8) we have (1&2 4 ) B #A &4.4 (mod ). Combnng (2.7) and (2.8), we get ( &3) nd(e &3 ) r &3 &1 B 4 1& #0 (mod ). Snce s rregular, certanly >5, and we get ( &3) nd(e &3 ) (r &3 &1)#0 (mod). Hence nd(e &3 )#0 (mod),.e., [E &3 Q]=1. Now we need the followng results from [S]: Lemma 2.2 [S]. of K. K(E 1 &3 ) s a nontrval, unramfed, abelan extenson Corollary. For any rncal deal (:) of K, [E &3 (:)]=1. Lemma 2.4 [He,. 434]. Let k # [5,..., l&2], and P k be any deal rme to (*) whose class belongs to C (k). Then [E &3P k ]=1. Snce [Q] generates C (3) by assumton, we can use Lemma 2.4 and the corollary of Lemma 2.2 to show that (E &3 ) 1 generates a trval extenson of Q[]. Ths wll contradct Lemma 2.2. Ths concludes the roof of the man theorem. Some Remarks. (1) By usng the other terms from (2.6), we get nd(e &t )#0 (mod) fort=5, and deendng on, some other t>3 as

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