## 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

$x^2 = m^2 - n^2, y^2 = 2m n, w = m^2 + n^2$

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 $$w$$.

Ben Lynn blynn@cs.stanford.edu 💡