Fermat's Last Theorem for exponent 4

Suppose a solution in positive integers to $x^4+y^4=z^2$ exists; choose one with $z$ minimal. If $d=\gcd(x,y)>1$, then $d^2\mid z$ and $(x/d)^4+(y/d)^4=(z/d^2)^2$ is smaller — contradiction. So $\gcd(x,y)=1$, $(x^2,y^2,z)$ is a primitive Pythagorean triple; say $x$ is odd. By Euclid's parametrization, coprime $m>n>0$ of opposite parity give $x^2=m^2-n^2$, $y^2=2mn$, $z=m^2+n^2$.

From $x^2+n^2=m^2$, the triple $(x,n,m)$ is itself primitive, so coprime $a>b>0$ give $x=a^2-b^2$, $n=2ab$, $m=a^2+b^2$. Then $y^2=4mab$, so $(y/2)^2=mab$ with $m,a,b$ pairwise coprime and product a perfect square, forcing each to be a square: $m=z_1^2$, $a=x_1^2$, $b=y_1^2$.

Substituting into $m=a^2+b^2$ gives $z_1^2=x_1^4+y_1^4$ — a new solution of the same equation, with $z_1\le z_1^2=m<m^2+n^2=z$: strictly smaller, contradicting minimality. No solution exists. For $x^4+y^4=z^4$, setting $Z=z^2$ would give a solution of $x^4+y^4=Z^2$, just ruled out.