## Problem 4

Determine the triples of non-zero natural numbers $(x, y, z)$ that satisfy the equation $x^{4}+y^{4}=2 z^{2}$, and the numbers $x$ and $y$ are coprime.
Solution (Prof. Lucian Petrescu, Tulcea):

We observe that the numbers $x$ and $y$ are odd.

We deduce that $z$ is also odd and is coprime with the number $x y$.

The equation in the statement transforms successively into $x^{8}+2 x^{4} y^{4}+y^{8}=4 z^{4}$, equivalent to $\left(x^{4}-y^{4}\right)^{2}=4 z^{4}-4 x^{4} y^{4}$, or $z^{4}-(x y)^{4}=\left(\frac{x^{4}-y^{4}}{2}\right)^{2}$.

We show that the equation $a^{4}-b^{4}=c^{2}(*)$, where $c \in \mathbb{N}$, the natural numbers $a$ and $b$ are odd, and $(a, b)=1$, holds only if $c=0$.

Assume that $c \neq 0$ and let $\left(a_{0}, b_{0}, c_{0}\right)$ be a solution to the equation ( $*$ ) with $a_{0}$ minimal.

From $a_{0}^{4}-b_{0}^{4}=c_{0}^{2}$, we deduce that there exist natural numbers $m>n>0$, of different parities, such that $a_{0}^{2}=m^{2}+n^{2}, b_{0}^{2}=m^{2}-n^{2}, c=2 m n$. Therefore, $m^{4}-n^{4}=(a b)^{2}$, meaning that the triplet ( $m, n, a b)$ is also a solution to (*). But since $a_{0}>m$, we obtain a contradiction with the minimality of $a_{0}$. It follows that $c=0$.

We deduce that $x=y$ and, since $(x, y)=1$, we obtain $(x, y, z)=(1,1,1)$.