]> Number Fields - Fermat's Last Theorem: n=4

## 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

${x}^{2}={m}^{2}-{n}^{2},{y}^{2}=2mn,{z}^{2}={m}^{2}+{n}^{2}$

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

$x={r}^{2}-{s}^{2},n=2rs,m={r}^{2}+{s}^{2}$

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 ${y}^{2}=4rsm$ 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 $w$.