6.26 For a positive integer $k$, there exist positive integers $n$ and $m$ such that $\frac{1}{n^{2}}+\frac{1}{m^{2}}=\frac{k}{n^{2}+m^{2}}$. Find all positive integers $k$.
(Hungarian Mathematical Olympiad, 1984)
[Solution] Removing the denominators from the known equation, we get
$$
\begin{array}{c}
m^{2}\left(n^{2}+m^{2}\right)+n^{2}\left(n^{2}+m^{2}\right)=k m^{2} n^{2}, \\
\left(m^{2}+n^{2}\right)^{2}=k m^{2} n^{2} .
\end{array}
$$

From this, we can see that $k$ is a perfect square. Let $k=j^{2}$, then
$$
m^{2}+n^{2}=j m n .
$$

If $(m, n)=d>1$, we can divide both sides of (1) by $d^{2}$, so we can assume that $m$ and $n$ are coprime.
Thus, from
$$
n \mid m^{2}+n^{2},
$$

we get
$$
n \mid m^{2},
$$

and hence
$$
n \mid m \text {. }
$$

Similarly, we also have
$$
m \mid n \text {. }
$$

Therefore,
$$
m=n \text {. }
$$

Given $(m, n)=1$, then $m=n=1$, so
$$
j=2 \text {. }
$$

Thus, $k=4$ is the only positive integer solution.