## Fermat’s Last Theorem for the case where the index is 4

Fermat’s Last Theorem, proved in 1995, is that the diophantine equation

$x^n + y^n = z^n$

has no solution in positive integers x, y, z when n > 2. Knowing the solution for n = 2 (see previous post) we can show that there is no solution for n = 4.  If

$x^4 + y^4 = z^4$   …………………………… (1)

has a solution x1, y1, z1 then

$x^4 + y^4 = z^2$   …………………………….   (2)

has a solution x1, y1, z12 and a solution x1/d, y1/d, z12/d2 where d = (x1, y1), the greatest common divisor of x1 and y1. This implies that (2) has a solution where (x, y) = 1. If we can show there is no such solution then by the contrapositive there is no solution to (1).

Suppose to the contrary there is such a solution x1, y1, z1. It is easy to see that these numbers have greatest common divisor 1. We use the method of infinite descent by showing that there exist x2, y2, z2 such that z2 < z1.

By the properties of Pythagorean Numbers (previous post) we know that

$z_{1} = a^{2} + b^2$  ;  $x_{1}^{2} = a^2 - b^2$  ;  $y_{1}^{2} = 2ab$

with a > b> 0, (a,b) = 1, and a, b of opposite parity.

x12 + b2 = a2 and it is not hard to show that that x1, b and a have the remaining properties of Pythagorean numbers. Thus

$x_1 = p^2 - q^2$  ;  $b = 2pq$  ;  and  $a = p^2 + q^2$

with (p,q) = 1, p > q > 0, and p,q of opposite parity. Consequently

$y_{1}^2 = 2ab = 4pq(p^2 + q^2)$

Furthermore p, q and p 2 + q2 have greatest common divisor one, so by the Unique Factorisation Theorem each is a square and we can write:

$p = x_{2}^2$  ;  $q = y_{2}^2$  ;  and $p^2 + q^2 = z_{2}^2$ and thus

$x_{2}^4 + y_{2}^4 = z_{2}^2$

with x2, y2, z2 > 0 and z1 = a2+b2 = (p2+q2)2+b2=z24+b2 > z2

To prove Fermat’s Last Theorem it suffices to prove it for every prime index.   If n > 2 is composite and has no prime factor other than 2 then it is a multiple of 4 and the Theorem is true for this index because if

$x^n + y^n = z^n$ then $\left( x^{n/4}\right)^4 + \left( y^{n/4} \right)^4 = \left( z^{n/4} \right)^4$

which is impossible.   If n is composite and has an odd prime factor p and if

$x^n + y^n = z^n$ then $\left( x^{n/p} \right) ^p + \left( y^{n/p} \right) ^p = \left( z^{n/p} \right) ^p$

thus by the contrapositive if the Theorem is true for p it is true for n.

A copy of this post in pdf form is here.