Fermat’s Last Theorem: n=4

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

\[x^2 = m^2 - n^2, y^2 = 2m n, z^2 = m^2 + n^2\]

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

\[ x = r^2 - s^2, n = 2r s, m = r^2 + s^2 \]

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 $w$.