An Elementary and Simple Proof of Fermat s Last Theorem

Transcription

1 A Elemetary ad Simple Proof of Fermat s Last Theorem Mie Wiler Faultät für Mathemati, Ruhr-Uiversität Bochum mie.wiler@ruhr-ui-bochum.de March 19, 2018 Abstract Fermat s Last Theorem states that the Diophatie equatio X + Y = Z has o o-trivial solutio for ay greater tha 2. I this paper we give a brief ad simple proof of the theorem usig oly elemetary methods. Itroductio The oly ow successful proof of Fermat s Last Theorem was give i 1994 by Adrew Wiles [5]. Ufortuately this proof cotais early hudred pages ad ca be uderstood i its etirety oly by some specialists. For this reaso ad relatig to Fermat s famous margial ote 1, may people (mostly amateurs are still looig for a shorter ad simpler proof based o elemetary methods. I this paper we preset such a proof. To prove Fermat s Last Theorem it suffices to prove it for the expoet 4 ad every odd prime expoet. A proof for the case = 4 has already bee give by Fermat himself. Therefore we give our proof oly for the prime expoets greater tha 2. First, we proof the correctess of the Diophatie idetity (z y = z y zy(z yλ,y,z with gcd(y, z, λ,y,z = 1 for ay odd prime. The we use the idetity z = axy b(x + y + c with a, b, c Z to covert the equatio x + y = z to the form bx +1 + b(z +1 y +1 = cx + ax(z +1 y +1. Fially, we show that x must divide (z +1 y +1. From this we deduce that x + y = z holds oly for the trivial solutios. 1 See [4] for the complete text.

2 The Proof Lemma 1. Let p, q be itegers with gcd(p, q = 1. The for ay odd prime there exists a iteger λ,p,q with gcd(p, q, λ,p,q = 1 such that (p q = p q pq(p qλ,p,q. (1 Proof. Accordig to the biomial theorem, we have ( (p q = p ( q 1 ( = p q + p ( q =1 1 = p q pq =1 1 2 = p q 1 pq(p q ( p 1 ( q 1 =1 (( 1 + ( 1 +1 p 2 ( q 1. (2 The term 1 (( 1 + ( 1 +1 assumes oly positive iteger values for ad odd prime. A proof ca be foud i Mamaai [2]. The OEIS referece for these values is A [3]. By substitutig λ,p,q = 2 =1 1 (( 1 + ( 1 +1 p 2 q 1 ito (2 we get (1. For = 3 we have λ 3,p,q = 1. For primes > 3 it follows from gcd(p, q = 1 ad the expasio of λ,p,q, give by p 3 ( 1 ( p 4 q pq 4 + q 3, (3 that gcd(p, q, λ,p,q = 1. Because if p divides λ,p,q the p q 3, ad if q divides λ,p,q the q p 3. This completes the proof. Lemma 2. Let a, b, c be itegers defied by the the idetity a = 1 2 x 2 (z +1 y +1, (4 b = 1 1 x 1 (z +1 y +1, (5 c = 1 x (z +1 y +1, (6 holds for all itegers x, y, z ad ay oegative iteger. z = axy b(x + y + c, (7 2

3 Proof. Multiplyig (4 by x gives Addig z y z y we get ax = 1 2 x 1 (z +1 y +1. Multiplyig by (z y yields Multiplyig (5 by x we have ax + z y z y = 1 1 x 1 (z +1 y +1 = b. ax(z y + z y = b(z y. (8 bx = 1 1 x (z +1 y +1. Addig z+1 y +1 z y we get bx + z+1 y +1 z y Multiplyig by (z y we obtai = 1 x (z +1 y +1 = c. bx(z y + z +1 y +1 = c(z y. (9 Now we ca prove the evidece of (7. Multiplyig (7 by (z y gives z (z y = axy(z y b(x + y(z y + c(z y. Applyig (8 o the right-had side we get z (z y = y ( b(z y z + y b(x + y(z y + c(z y = bx(z y yz + y +1 + c(z y. Applyig (9 o the right-had side yields z (z y = bx(z y yz + y +1 + bx(z y + z +1 y +1 = z +1 yz. We obtai a true statemet, which completes the proof. Lemma 3. Let x, y, z be ozero itegers with gcd(x, y, z = 1 ad let be a odd prime. Accordig to Lemma 2, if x ad (z y x, we have gcd(x, b = 1. Proof. Let p x be a prime factor of (z y ad x. For = 0,..., 2 we have z +1 y +1 z y = z i y i. i=0 3

4 Thus it follows from (5 that oly z y z y decides whether p x divides b, because its the oly term i the sum without the factor x. Applyig Lemma 1 with p = z, q = y, we obtai z y z y = (z y 1 + zyλ,z,y, (10 with gcd(z, y, λ,z,y = 1. Because x, y, z are pairwise relatively prime, exactly oe of these itegers is eve. If x is eve the y, z are odd, so (z y is eve. Applyig Lemma 1 it follows from (3 that λ,z,y is odd, because the sum cosists of a odd umber of terms, where the umber of eve coefficiets is also eve ad the umber of odd coefficiets is also odd. If x is odd the y, z have differet parity, so (z y is odd. Applyig Lemma 1 it follows from (3 that λ,z,y is odd, because each term i the sum is eve except (z 3 +y 3 which is odd. From gcd(z, y = 1 we have gcd(z, y, z y = 1, hece gcd(z y, λ,z,y = 1, so gcd(z y, zyλ,z,y = 1. It follows with (z y x, x ad x ±1 that p x zyλ,z,y. Hece p x does ot divide the right-had side of (10, which completes the proof. Theorem 4. The Diophatie equatio X + Y = Z has o o-trivial solutio for ay odd prime umber. Proof. We assume that x, y, z are ozero itegers ad is a odd prime such that x + y = z. (11 It suffices to cosider oly solutios (x, y, z with gcd(x, y, z = 1. Hece x, y, z are pairwise relatively prime ad exactly oe of these itegers is eve. Applyig Lemma 1 with p = x + y, q = z, we have (x + y z = (x + y z (x + yz(x + y zλ,x+y,z. Applyig Lemma 1 with p = x, q = y, o the right-had side gives (x + y z = x + y + xy(x + yλ,x, y z (x + yz(x + y zλ,x+y,z. Applyig (11 o the right-had side we obtai that is (x + y z = xy(x + yλ,x, y (x + yz(x + y zλ,x+y,z, (x + y z = (x + y ( xyλ,x, y z(x + y zλ,x+y,z. (12 Because is a odd prime we coclude from (12 that (x+y z, hece (x+y z. From gcd(x, y, z = 1 it follows that divides oe ad oly oe of the itegers x, y, z. Applyig (11 o the right-had side of (8 we obtai hece ax(z y + x = b(z y, x = (b ax(z y. (13 4

5 From (9 we have Combiig (13 with (14 yields that is z +1 y +1 = (c bx(z y. (14 x (c bx = (z +1 y +1 (b ax, bx +1 + b(z +1 y +1 = cx + ax(z +1 y +1. (15 Now we assume that x ±1 ad x. Applyig Lemma 1 with p = z, q = y, it follows from (11 that (z y x. From gcd(x, y, z = 1 ad gcd(z, y, λ,z,y = 1 we have gcd(z y, λ,z,y = 1, ad so (z y x. By Lemma (3 we have gcd(x, b = 1. Hece, dividig (15 by x we obtai bx + b z+1 y +1 x = cx 1 + a(z +1 y +1, which gives x (z +1 y +1. Hece x divides the right-had side, so bx 1 + b z+1 y +1 x 2 = cx 2 + a z+1 y +1, x ad cosequetly x 2 (z +1 y +1. Thus it follows that (15 ca be divide times by x, which yields Multiplyig (11 by z gives Subtractig y +1 we obtai Dividig by x yields bx + b z+1 y +1 x = c + a z+1 y +1 x 1. (16 zx + zy = z +1. zx + y (z y = z +1 y +1. z + y (z y x = z+1 y +1 x. (17 From (16 we coclude that x (z +1 y +1. Hece, from (17 it follows with gcd(x, y, z = 1 that x (z y. This clearly forces with (z y x that x = ±1, which is a cotradictio to our assumptio x ±1. Hece, a o-trivial solutio with x ca ot exists. I the case x we ca iterchage x ad y, which completes the proof. Remar 5. The terms from (4 (6 represet special cases of the triomial expasio of (x + y + z with the peculiarity that all triomial coefficiets give by ( i,j, =! i! j!! were set equal to 1. Let x, y, z be itegers, the for ay oegative iteger we have 1 x (z +1 y +1 = x i y j z, i+j+= where i, j, are all oegative itegers such that i + j + =. 5

6 Remar 6. We ca rewrite the terms from (4 (6 as fractios. From (4 we obtai a = z 2 x 2 z y 2 x 2 y = z z 1 x 1 y z x y 1 x 1 y x = x (z y + y (x z + z (y x (z y(x z(x y I a similar way, we may show that = x (z y + y (x z + z (y x x 2 (z y + y 2 (x z + z 2 (y x. b = x+1 (z y + y +1 (x z + z +1 (y x x 2 (z y + y 2 (x z + z 2, (y x c = x+2 (z y + y +2 (x z + z +2 (y x x 2 (z y + y 2 (x z + z 2. (y x Table 1 gives a overview o all possible iteger values for a, b, c from (4 (6 for ay oegative iteger. a b c Z 2 1 Z N odd 3 Z N Z eve 4 N Z N Table 1: Possible values for a, b, c. Acowledgemets The author wishes to express his thas to Adreas Fillipi. Without his cotributio i a mathematics forum, I probably would ever have wored o this topic agai [1]. Refereces [1] Fillipi, Adreas: Beitrag im Forum Elemetare Zahletheorie, February 9, 2015, (tiyurl.com/y8jh4dvr [2] K. Mamaai, F. Rusey, New roses: simple symmetric Ve diagrams with 11 ad 13 curves, Disc. Comp. Geom., 52 (2014, pp , Lemma 2. 6

7 [3] The O-Lie Ecyclopedia of Iteger Sequeces: T(, is the umber of -poits o the left side of a crosscut of simple symmetric -Ve diagram, A [4] Wiipedia, Fermat s Last Theorem, 2.2 Fermat s cojecture. (tiyurl.com/y9hwwdb5 [5] Wiles, Adrew: Modular Elliptic Curves ad Fermat s last theorem, Aals of Mathematics 142 (1995, pp