Simple and analytical proof of Fermat's last theorem for n=3

Abstract

It is well known that Fermat's last Theorem, in general, is extremely difficult to prove although the meaning of the theorem is very simple. It is surprising that the proof of theorem for n = 3 ,the smallest, corresponding number, given by Leonard Euler, which has been recommended for amateurs[!], is not only difficult but also has a gap in the proof. Paulo Rebenboim claims[1] that he has patched up Euler's proof, which is very difficult to understand, however. It was shown[2] that the parametric solution for the x,y,z in the equation z3 = y3 +x3,(x,y) = 1 could be obtained easily with one necessary condition that must be satisfied by the parameters. Fairly simple analytical proof of the Fermat's last theorem for n = 3 was given [2] using this necessary condition. In this contribution very much simpler proof is given ,which is very suitable for amateurs. Fermat's last theorem for n = 3 can be stated as the equation z3 = y3 + x3, (x,y) ·= 1. (1) has no non-zero integral solution for (x,y,z). If we assume a non-zero( xyz * 0) solution for (x,y,z) , then one of (x,y,z) is divisible by 3 . Since we can assume for (1) for negative integers, without loss of generality one can assume that y is divisible by 3 . Then if y = 3fJ ay, the parametric solution of (1), can be expressed as X = 3fJ a()o + 03 y = 3fJ a()o + 33fJ-1a 3 z = 33fJ-1 a3 + 3fJ a()o + 83 and a necessary condition satisfied by the parameters is ()3 -83 -2.3fJ aoB -33[3-1a 3 =0 In this equation ,() is a factor of z, 8 is a factor of x and r = B5 + 32P-I a2 Proof of the Fermat's last theorem for n = 3 (a) (b) (c) (d) Expressing 33/3-I a3 as 33/3-3 a3 + 8.33/3-3 and substituting e = 3fJ-1 g + o m (d),one gets (g- 2a)(o2 +3fl-1go+32fl-3(g2 +2ag+4a2 ))=32fl-3a3 (2) the condition,()= 3/3-l g + 8 is due to a simple lemma used in [2] from which it follows that f3 > 1 .It is easy to deduce that g-2a is divisible by 3 2/3-3 since (3, o) = 1 and fJ > 1 . If a= ± 1 , then ,g = ±2+32P-3>0. Now, (82 + 3/3-l g8 + 32f3-\g2 +. 2ag + 4a2 ) = ±1 is never satisfied since (2) can be expressed as 3/3-1 2 (8 +-- g )2 +32P-3(L+2g +4)= ±1 2 4 (3) (4) If a* ±1, we deduces from (2) that g-2a= 3 213-3s 3 where s* 1 and s is a factor of a . This is because factor of g cannot be a factor of both 8 and a since B = 3/3-l g + 8 and a is a factor of y . Hence, (5) where a= sq, (s, q) = 1. Now, this quadratic equation ing must be satisfied by 2a +3 213-3s 3. If the roots of this quadratic are a1 and a 2 =2a + 32/3-3 s3, then 82 + 4a2 .32/3-3 _ q3 2a.32f3-3 + 3p-t8 ala 2 = 3213_3 , a1 + a 2 =- 3213_3 It is easy to obtain from these two relations that (6) Hence, 2a + 3213-3s 3 is a factor of 82 + 4.a2 .32/3-3- q3 . In other words, 2a +3 2/3-3s 3 is a factor of the expression (2a)3 -(2a)2 .8s33 2/3-3 -882s3 or, -32/3-3s 3 is an integral root of the equation x3 -8s33213-3x2 -882s3= 0 (7) (8) This means that 32/3-3s 3 is a factor of 882. s3 which contradicts (3, 8)= 1 , that is (3,x) = 1, and the proof of theorem is complete. It should be emphasized that the necessary condition needed for our proof can be obtained without obtaining the parametric solution of (1 ), making the proof given here much shorter.