Analytical proof of Fermat's last theorem for n=4

Abstract

Fermat's last theorem for n = 4 is usually proved [1] using the famous mathematical tool of the method of infinite descent of F ermat. In this contribution, it will be shown that the parametric solution of the polynomial equation d4 = e4 + g4, (e, g) = 1 can be obtained using a simple mathematical technique and thereby the proof of the theorem can be done, without depending on the sophisticated structure of primitive Pythagorean triples of Fermat[1] given by X= 2lm, y =!2-m2 , z = !2 + m2 ' where l >m> 0 and l, m are of opposite parity. The main objective of this contribution is to introduce a new simple mathematical technique which may be very useful in some other problems as well. If the equation (1) has a non-trivial integral solution for (x, y, z) , then one of e, g is even and we can assume that d, e, g are positive. If gis even, (d2-g2)(d2+g2)=e4 and terms in the brackets are eo-prime and hence , one writes d2 +g2 =x4 d2-g2 =y4 (2a) (2b) From these two equations, we get 2d2 = x4 + y4. (2c) Therefore (x2 -d)(x2+ d)=( d-y2 )(d + y2) and it is easy to deduce that terms in the brackets on the left-hand side or on the right-hand side of this equation may have only factor 2 in common since all numbers are odd and x, d, y are eo-prime to one another. In the following, a new simple mathematical technique is used to obtain the parametric solution for x, y ,d, g from this single equation. If x2 -d = d-y2, 2d = x2 + y2 and therefore 4d2 = x4 + y4 + 2x2 y2 , which means d2 = x2y2, and it leads to a contradiction since (d, e) = 1. Similarly we can easily show thatx2-d:t=d+y2. Now, let (d-y2)= a (x2-d) ,to obtain x2-d = ba-1(d-y2), b where(a, b) = l.Then k (x2 +d )= (d+ y2), x2+ d = ab-1( d + y2) .Now, let us form the a following two simultaneous equations, to obtain, x2-d =ba-1(d-y2) (a) x2 + d = ab-1(d + y2) (b) (3) Proceedings of the Annual Research Symposium 2008- Faculty of Graduate Studies University of Kelaniya (4) Since ( d, y) = 1 , b 2 + a 2 = dk , where k has to be determined. Then , one easily obtains 2 2ab+b2-a2 a2+b2 . 2ab = a2 -b2 + y2 k , y = , d = .Now , from(3), It follows that k k 2abx2 = (a2 +b2)d +(a2 -b2)y2 = (a2 +b2)2 +(a2 -b2)(2ab+b 2 -a2) k (5) It is clear from (5) that a and b cannot be of opposite parity since then k2 x2 y2 cannot be either odd or even. Hence a and b are both odd. and therefore k2 = 4 or 41 k2. a2 +b2 Thereforex 2y2 =(4a2b2 -(a2 -b2)2)14 =e2, d = , ab (a2 -b2)=g2 as given 2 below , which is the parametric solution of the equation (1 ), where a, bare parameters. 2 2ab+a2-b2-a2-b2 2a (a-b) 2 2 Now, x -d = = and k is a factor of a +b and if k k it is a factor of a-b , one deduces that k is 2 or a factor of a or b. Since (a, b)= 1, we conclude that k = 2 . Therefore x2 y2 = a2 b2 -( a2 -b2 )2 /4 Since (x2-y2)(x2 + y2) = x4-y4 = 2g2, which follows from(2a),(2b), it IS easy to deduce (6) Therefore a, b, (a2-b2) should be perfect squares. Now, if a=r2, b =s2, then r4 -s4 = t2 for some integers r, s, t. The famous and the only theorem that Fermat has proved is that there are no integers r, s, t satisfying r4-s4 = t2. Hence the Fermat's last theorem for n = 4 can be deduced. It is quite interesting that applying the mathematical technique used in this contribution ,we have shown[2] very easily that the equation r4-s4 = t2 has no non- trivial integral solution for r,s,t ,and then the Fermat's last theorem for n = 4 follows at once[1].