Analytical Proof of Fermat's Last Theorem for n = 3

Abstract

1. Introduction It is well known that the proof of fermat's Last Theorem, in generaL is extremely difficult. It is surprising that the proof of theorem for n = 3, the smallest corresponding number, given by Leonard Euler, which is supposed to be the simplest, is also difficult and erroneous. Paulo Rebenboin claims that he has patched up [1] the Euler's proof, which is very difficult to understand. however. In this article we present a simple and short proof of the Fermat's last theorem. Fermat's Last Theorem The equation z" = y" + x". (x, y) = 1 has no nontrivial integral solutions (x,y,z) for any prime n ;::: 3 . 2. Proof of the Fermat's last theorem for n = 3 In the following, the parametric solution to the problem based on very simple three lemmas is given. 2.1 Lemma If a 3 - b3 is divisible by Y' (p * 0) and (a, 3) = 1 = (b, 3), then (a-b) is divisible by 3~'- 1 and p;::: 2. This lemma can be easily proved substituting a- b = k in a3 - b3 and we assume it without proof. 2.2 Lemma If the equation (1) has a non trivial integral solution (x,y,z), then one of x,y,zis divisible by 3. Proof of this lemma is also simple and we assumed it without proof. Now. (1) takes the form (2) 2.3 Lemma There are two integers a and f3 such that z-x=331Ha3 , (3,a)=l. Proof of this lemma is exactly the same as in the case of analytic solution of Pythagoras' theorem [2] and let us assume it without proof. Now (2) takes the form .:3 = 33fJ a 3ry 3 + x3 , (3, z) = 1 (3) From the equation 3 y s (y + s) = x 3 - s 3 , it follows that s should be of the form 6 3 , where (6, 3) = 1 since s divides x and (s, y) = 1. Also note also that 6 3 = z- y. Then the above equation becomes 3 y63 (y +53)= x3 - 69 Now it is clear that x -15 is divisible by three. Let us consider the expression x + y- z. x + y- z = x- (z- y) = x -153 . Now consider the original equation z 3 = y 3 + x 3 , (x, y) = 1 in the form that z 3 = (x + y) (Cx + y) 2 - 3xy ). It is clear that x + y and the term, (x + y) 2 - 3xy are co-prime and therefore x + y = () 3 , where z = er; and ((), r;) = 1. Now again X+ y- z = () 3 - er; = ()(() 2 - r;) and therefore X -15 3 is divisible (). x + y- z = y- (z- x) = 3fJ ary- 33!3 a 3 = 3fJ a(ry- 32/3-l a 2 ) and therefore x -15 3 is divisible by 3 fJ a. x is divisible by 15, which follows from ( 4) and therefore x -15 3 is divisible by 3fJ a()l5. Now consider (4) m the form 3f3+ 1aryl5 3()r;=(x-15 3 Xx 2 +xl53 +15 9 ). From which one understands that x -15 3 = 3fJ a()l5 and since z- x = 3 3 ~-l a 3 . x = 3fJ aBI5 + 153 (a) y = YaBI5 + 33.LJ-l a 3 (b) z = 33/3-l a 3 + 3fJ aBI5 + 15 3 (c) In addition to this, we have x + y = () 3 and ('7- 32/3-l a 2 ) = ()15, and therefore substituting for 17 in y , we get ()3 -15 3 - 2.3.LJ aBI5- 33.LJ-t a 3 = 0 (d) Therefore by lemma (1), B-15 should be divisible 3tH. Expressing 33fJ- 1a 3 as 8.3 3!3-3 a 3 + 33!3-3 a 3 and () = 3.LJ-t g + 15 ,we obtain from (d) that (g-2a)(152 +3P-1gi5+3 2P-3 (g 2 +2ag+4a2 ))=3 2P- 3 a 3 (5) If X = (g- 2a ), Y = (15 2 + 3/3-l gl5 + 32/3-J (g 2 + 2ag )+ 4a 2 ), then it 1s clear that (3,Y) = 1, Now we prove the Fermat's last theorem for n = 3, showing that (d) is never satisfied. If 3,Li-l 2 a=1, g=2+32.LJ-J and Y=1=(15+--g-)2 +32.LJ-3(_[__+2g+4) which is never 2 4 satisfied. Similarly , the proof of the theorem follows when a = -1 smce f3 ~ 2 . If a 1:- 1, g = 2a. + 32.LJ-J q3 and () = 2a3P-I + (3 3.LJ-4 p 3 + 15). The equation (d) is of the form () 3 -3(2.3P-1a)I5B-8.33,u-3 a 3 -(33P-3 a 3 +15 3)=0 (6) and is also of the form. x 3 - 3.u.vx- u3 - v3 = 0 and therefore we can make use of the well known method of Tatagliya and Cardon (see[3] ).Then u 3 must be a solution of the quadratic x 2 + Gx - H 3 = 0 and the roots of (d) for ()are u + v, u UJ + v UJ 2 , u UJ 2 + v UJ , where UJ is the cube root of unity. It can be easily shown that this occurs only if u = 0. which gives a = 0 and this corresponds to the trivial solution x.y.z = 0 .Hence the proof of the theorem.