4-179 For a given positive integer $m$, find all positive integer triples $(n, x, y)$, where $m, n$ are coprime, and satisfy
$$
\left(x^{2}+y^{2}\right)^{m}=(x y)^{n} .
$$
[Solution] We will prove that when $m$ is odd, the equation has no solution; when $m$ is even, the only solution is
$$
(n, x, y)=\left(m+1,2^{\frac{m}{2}}, 2^{\frac{m}{2}}\right) \text {. }
$$

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

thus $\quad n>m$.
Let the highest powers of a prime $p$ in $x, y$ be $a, b$ respectively. Then the highest power of $p$ in $(x y)^{n}$ is $n(a+b)$.

If $a < b$, then the highest power of $p$ in $(x y)^{n}$ is $n(a+b)$, while the highest power of $p$ in $(2 x y)^{m}$ is $m(a+b+1)$. Since $n > m$, this leads to a contradiction.
Similarly, if $a > b$, we can also derive a contradiction.
Therefore, for all primes $p$, we have $a=b$. Hence,
$$
x=y \text {. }
$$

Thus, the original equation reduces to
$$
\left(2 x^{2}\right)^{m}=x^{2 n},
$$

which simplifies to
$$
x^{2(n-m)}=2^{m} \text {. }
$$

From the above equation, we know that $x$ is an integer power of 2. Let $x=2^{k}$, then
$$
\begin{array}{l}
2(n-m) k=m, \\
2 n k=m(2 k+1) .
\end{array}
$$

Since $\quad(m, n)=1,(2 k, 2 k+1)=1$,
we have $m=2 k, n=2 k+1$.
This means that $m$ must be even, $n=m+1, x=2^{\frac{m}{2}}, y=2^{\frac{m}{2}}$. Therefore, when $m$ is odd, the equation has no solution; when $m$ is even, the only solution is
$$
(n, x, y)=\left(m+1,2^{\frac{m}{2}}, 2^{\frac{m}{2}}\right) .
$$