Almost Pythagoras

Mathologer made a video on this :)

Nice solution! Can you provide a proof that $x^4 + y^4 = z^2$ has no solutions for everybody's enjoyment?

Fermat, famous for his unproven claim that $x^n + y^n = z^n$ has no non-trivial solutions for $n \geq 3$ , did actually provide a prove for this.

Suppose, w.l.o.g., that $(x,y,z)$ have no common factors and $y$ is even. Then $(x^2, y^2, z)$ are a primitive Pythagorean triple. This means that $x^2 = p^2 - q^2$ , $y^2 = 2pq$ and $z = p^2 + q^2$ , for some odd $p$ and even $q$ .

But now we have $x^2 + q^2 = p^2$ , showing that $(x, q, p)$ are also a Pythagorean triple (which is actually primitive). Let's say $x = u^2 - v^2$ , $q = 2uv$ and $p = u^2 + v^2$ . Now $y^2 = 2pq = 2\cdot (u^2 + v^2)\cdot 2uv = 4uv(u^2 + v^2)$ must be a square. However, $u$ , $v$ and $u^2+v^2$ are coprime, so that they must all be squares; say, $u = \xi^2$ , $v = \eta^2$ , and $u^2 + v^2 = \zeta^2$ , with $\xi, \eta, \zeta$ coprime. But now we have a triple $\xi^4 + \eta^4 = \zeta^2,$ We now have a recipe to replace any coprime triple $(x,y,z)$ such that $x^4 + y^4 = z^2$ by a strictly smaller coprime triple $(\xi,\eta,\zeta)$ with the same property. This process could be repeated indefinitely, $(x,y,z) \mapsto (\xi,\eta,\zeta) \mapsto \dots$ However, such an infinitely descending sequence of positive integers does not exist, proving that there cannot be any triple $(x,y,z)$ with the desired property.