A-3. For a given positive integer $m$, find all positive integer triples $(n, x, y)$, where $m, n$ are coprime, and satisfy
$$
\left(x^{2}+y^{2}\right)^{m}=(x y)^{n} .
$$
When $m$ is odd, the equation has no solution. When $m$ is even, the only solution is $(n, x, y)=\left(m+1,2^{\frac{m}{2}}, 2^{\frac{m}{2}}\right)$. We will prove this below.

If $(n, x, y)$ is a solution to the equation, then by the arithmetic-geometric mean inequality, we have $(x y)^{n}=\left(x^{2}+y^{2}\right)^{n} \geqslant(2 x y)^{n}$. Therefore, $n>m$.

Let $p$ be a prime number, and let $a, b$ be the highest powers of $p$ in $x, y$ respectively. Then $(x y)^{n}$ contains the highest power of $p$ as $(a+b)^{n}$. If $a<b$, then $\left(x^{2}+y^{2}\right)=$ contains the highest power of $p$ as $2 a n x$, which implies $x=y$. Thus, the equation becomes $\left(2 x^{2}\right)^{m}=x^{20}$ or equivalently $x^{2(n-m)}=2^{m}$. Hence, $x$ must be an integer power of 2. Assume $x=2^{n}$, then $2(n-m) a=m$ or $2 a n=(2 a+1) m$. Since
$$
\operatorname{gcd}(m, n)=\operatorname{gcd}(2 a, 2 a+1)=1,
$$

we must have $m=2 a$ and $n=2 a+1$. Therefore, $m$ must be even, and the solution is as stated.
($\operatorname{gcd}(m, n)$ refers to the greatest common divisor of $m, n$—translator's note)