On the Diophantine equation x 2 2 ˆ y n

Transcription

1 Arch. Math. 74 (000) 50± /00/ $.70/0 Birkhäuser Verag, Base, 000 Archiv der Mathematik O the Diohatie equatio x ˆ y By B. SURY Abstract. We give a eemetary roof of the fact that the oy soutios of the Diohatie equatio x ˆ y for > 1 are x; y; ˆ 5;;. Itroductio. It is kow due to T. Nage ( Š) that the equatio of the tite has oy the oe soutio x; y ˆ 5; for > 1. But, Nages roof is ot eemetary as it uses a dee resut of K. Maher o biary quadratic forms. More recety, J. H. E. Coh ( 1Š) ivestigated the equatios x k ˆ y for odd k. However, he merey refers to Nages roof whe k ˆ 1 ad his (eemetary) roof for higher k does ot work for k ˆ 1. I the iterature, there are essetiay two kids of methods used to sove such Diohatie equatios. Either oe uses trascedeta umber-theoretic techiques or a cometey eemetary techique maiuatig cogrueces. The urose of this ote is to give a eemetary roof of Nages resut. We sha use a oyomia idetity i the roof as the usua eemetary aroach turs out to be iadequate. Some stadard reductios. Suose x; y; satisfy the equatio of the tite. It easiy foows that x; y; must a be odd. Writig the equatio as y ˆ x i x i ad otig that Q has cass umber 1, it foows that oe must have x i ˆ a where a ˆ m ir for some itegers r; m. The, x i ˆ a ad subtractig, oe has a a ˆ i. Oe ca rewrite this as 1 =Š m 1 r 1 ˆ 1: 1 This gives r ˆ1 ad, readig moduo, aso that both ; m are odd. We ote that, coversey, if there are itegers m; ; r satisfyig the above equatio, the the itegers x; y defied by them give a soutio of the Diohatie equatio of the tite. For arbitrary itegers m; ; r, et us write 1 =Š a m; ; r ˆ m 1 r 1 : 1 Note that a m; ; r ˆa a a a where a ˆ m ir. We eed to sove a m; ; r ˆ1 for m; ; r. We start with some usefu observatios: Mathematics Subject Cassificatio (1991): 11D41.

2 Vo. 74, 000 O the Diohatie equatio x ˆ y 51 Lemma. (i) If a m; ; r ˆ 1, the r ˆ 1. (ii) a m; ; 1 ĵ 1 for ay m;. (iii) If a m; ; 1 ˆ 1, the a m; d; 1 ˆ 1 for every dj. (iv) For ay m, a :ˆ a m; ; 1 satisfies the foowig recursio i : a 1 ˆ ma m a 1. (v) If ˆ ad a m; ; 1 ˆ 1, the m ˆ 1 i.e., the oy soutios of x ˆ y are x; y ˆ 5;. (vi) If a m; ; 1 ˆ 1, the mod 4. I this case, if ĵ, ad s is the highest ower of dividig, the m 1 s mod s 1. I articuar, m ĵ 1 if ĵ i.e., x ˆ has o soutios if ĵ 1;. P roof. (i) As we oticed above, if a m; ; r ˆ 1, the m; are odd ad r ˆ 1. Now, m 1 mod 4; so m 1 ˆ m 1 = 1 ad m 1 mod 4. Further, we have 1 = 1 ˆ a m; ; r ˆ m 1 r 1 1 m 1 r m r r r mod 4: 1 As 1 1 mod 4, oe has 1 ˆ 1 mod 4. Thus, 1 ˆ a m; ; r r r 1 mod 4: If 1 mod 4, oe has the 1 r mod 4. If 1 mod 4, oe has 1 r 6r mod 4. I either case, r ĵ 1. As r ˆ 1, we get r ˆ 1. (ii) If a m; ; 1 ˆ 1, the evidety a m; ; 1 ˆ 1 from the very defiitio of a m; ; r. This cotradicts (i). a m; ; 1 (iii) Now a m; d; 1 ˆ a a a d is i Q \ Z ˆ Z. So, a m; d; 1 ja m; ; 1 if dj. This roves a d our assertio i view of (ii). (iv) Now a ˆ a a for ay where a ˆ m i. So, oe has a a a a a 1 ˆ a 1 a 1 ˆ a a a aa a 1 ˆ a a a aa aa aa a 1 ˆ a a a aa a 1 a 1 a a a ˆ a a a a aa a 1 a 1 ˆ m a a m a 1 a 1 : Therefore, a 1 ˆ ma m a 1. (v) Is obvious as 1 = 1 ˆ a m; ; 1 ˆ m 1 1 reduces to 1 ˆ m i.e., m ˆ 1 the.

3 5 B. SURY ARCH. MATH. (vi) Reca that m; must be odd as a m; ; 1 ˆ 1. Now, if 1 mod t, the k ˆ k 1 0 mod t 1 k 1 k 1 k k 1 for k ^ as k is eve i.e., the ower of dividig k is ess tha k. k Suose a is the highest ower of dividig 1 i.e., 1 a mod a 1. Usig the above fact ad writig () moduo a 1, oe gets 1 = 1ˆ m 1 m 1 m 4 m 5 mod a 1 : Let us ook at the three terms oe at a time. As m is odd, we have m a ˆ m a 1 1 mod a 1 ad as 1 mod a, we aso have m 1 1 mod a 1. Usig 1 a mod a 1, the first term is the m 1 1 a mod a 1 : Now, the secod term is m 1 ˆ m. Let us substitute 1 a mod a 1 ad write 1 for the iverse of moduo a 1. The, the secod term is 1 1 a a a 1 m mod a 1 : Usig further the fact that a t a mod a 1 if t is odd, the secod term is see to be a moduo a 1. Fiay, the third term is 1 4 m a a a 1 a a m 5 mod a a a a 1 a a m 5 mod a 1 1 a a a 1 a 1 1 a m 5 mod a 1 : If a > 1, a 1 1 is odd, ad the third term is a mod a 1. Therefore, if a > 1, the sum of the three terms is 1 1 a a a ˆ 1 a mod a 1 which is absurd. Hece, we cocude that if a m; ; 1 ˆ 1, the a must be 1 i.e., the statemet 1 a mod a 1 simy becomes mod 4. This was the first assertio i (vi). Fiay, if mod s, the k ˆ k mod s 1 k 1 k 1 k k 1 k k for k ^ as the ower of dividig the deomiator is at the most k.

4 Vo. 74, 000 O the Diohatie equatio x ˆ y 5 Takig for s the maximum ossibe vaue, we have s mod s 1. As before, we rewrite () moduo s 1 usig the above cogruece to get: 1 = 1ˆ a m; ; 1 ˆ m 1 m m m 5 mod s 1 : Substitutig s mod s 1, this becomes 1 s m 1 1 s s s s s s 1 s s 1 m 5 mod a 1 : Agai, we ook at the three terms oe by oe. Sice m s 1 mod s 1, we have m 1 mod s 1 i.e., m 1 m mod s 1. Usig this as before with the evidet observatio that s t s mod s 1 if t is odd, the first term is s m 1 s m m s mod s 1 : Simiary, the secod term is s mod s 1. The third term is 15 1 s s s 1 s s 1 m 5 ˆ 15 1 s s 1 1 s 1 s s 1 m 5 ˆ t s s mod s 1 because t is odd. We have used the fact that s > 1 (i.e., that mod 4) which we roved aready. So, the sum of the three terms is 1 ˆ a m; ; 1 m s mod s 1 : Thus, oe obtais m 1 s mod s 1. I other words, m 1 1 s s mod s 1 : The roof of the emma is comete. We sha use aother idetity to augmet some of the iformatio give by the emma ad comete the soutio of the Diohatie equatio. It is ot cear to us as to how to use arts (iii) ad (iv) of the emma. A oyomia idetity to comete the roof. As a off-shoot of a agebro-geometric questio about embeddigs of the affie ie i -sace, the curious oyomia idetity k=š 1 k Y Y k ˆ k d Y k d dˆ0 was oticed i ( Š). This turs out to suit our reset urose we. Take for ; Y the comex umbers a ad a, ad k ˆ 1 with odd. As a ˆ m i, we ote that aa ˆ m ad a a ˆ m. Now, we have a m; ; 1 ˆ a a a a ˆ P 1 a d a 1 d : dˆ0

5 54 B. SURY ARCH. MATH. Usig the idetity, we get therefore A a m; ; 1 ˆ 1 = 1 1 m m 1 : Let us observe i assig that we have obtaied for ay odd a idetity i Z TŠ: 1 = = T T 1 ˆ r T r 1 : r 1 To comete the soutio of the Diohatie equatio, we start with the assumtio that a m; ; 1 ˆ 1. We may aso take > because we have aready deat with ˆ i the emma ad we kow that is odd. We sha write a istead of a m; ; 1 for simicity. By the emma (vi), oe has a iteger a ^ such that a mod a 1 ad m 1 a mod a 1. Let us comute a mod a 1 usig (A). Note that m k k or k a mod a 1 accordig as k is eve or odd. Therefore, mod a 1, oe has 1 ˆ a 1 m 1 1 a m 1 eve odd 1 = 1 1 m 1 odd rˆ0 1 a m 1 mod a 1 : Sice ^ mod 4 by emma (vi), we kow that 1 = is odd. Aso, uess ˆ 1 =, we get a m 1 0 mod a 1. Therefore, we have 1 = 1 ˆ a 1 1 m 1 a mod a 1 where a is the term corresodig to ˆ 1 = i the secod sum. Fiay, emma (vi) gives m 1 a mod a 1 which imies that m r r mod a 1 for each r > 0. Thus, Writig b ˆ 1 = 1 ˆ a 1 = a mod a 1 : 1 1, we get B b 1 a mod a 1 : Now b ˆ b b where b ˆ 1 i. Let us write ˆ a b with b odd. The biomia b b exasio gives b ab ˆ 1 a bb a 1 for some Z Š. Sice b ˆ b 6, we get b ˆ b b b b ˆ 1 a 1 m for some m Z Š. As m Z \ Q ˆ Z, we fiay obtai b 1 mod a 1. This cotradicts (B). We have therefore roved Nages resut:

6 Vo. 74, 000 O the Diohatie equatio x ˆ y 55 Theorem. The Diohatie equatio x ˆ y, > 1 has oy the soutios x; y; ˆ 5; ;. Ackowedgemets. We woud ike to thak the referee for suggestios to imrove the exositio ad to carify some oits. Refereces [1] J. H. E. COHN, The Diohatie equatio x k ˆ y. Arch. Math. 59, 41 ± 44 (199). [] T. NAGELL, Veragemeierug eies Fermatsche Satzes. Arch. Math. 5, 15 ± 159 (1954). [] B. SURY, A curious oyomia idetity. Nieuw Arch. Wisk. 11, 9 ± 96 (199). Aschrift des Autors: B. Sury Stat-Math Divisio Idia Statistica Istitute 8th Mie, Mysore Road R.V. Coege P.O., Bagaore , Idia Eigegage am