We prove Fermat’s Last Theorem for this case by showing \(x^4 + y^4 = w^2\) has no solutions in the positive integers.

Suppose there is a solution. Then let \(x,y,w\) be a solution with the smallest possible \(w\). First note \(x^2, y^2, w\) form a Pythagorean triple. Without loss of generality assume \(x\) is odd, so write

for coprime \(m,n\) that are not both odd.

Then the first equation implies that \(x, n, m\) also form a Pythagorean triple with \(x\) odd, so we may write

for coprime integers \(r,s\) that are not both odd.

The last of these three equations implies \(r,s,m\) are pairwise coprime (otherwise \(r,s\) could not be coprime) and from \(y^2 = 4 r s m\) we deduce that \(r = a^2, s = b^2, m = c^2\) for some integers \(a,b,c\).

But substituting these in the equation for \(m\) implies that \(a^4 + b^4 = c^2\), contradicting the minimality of \(w\).