19.2.3 * For a positive integer $k$, there exist positive integers $n, m$, such that $\frac{1}{n^{2}}+\frac{1}{m^{2}}=\frac{k}{n^{2}+m^{2}}$, find all such $k$.
From the original equation, we get $\left(\frac{n}{m}+\frac{m}{n}\right)^{2}=k$. Without loss of generality, assume $m$ and $n$ are coprime (otherwise, factor out the common divisor). Since $\frac{n}{m}+\frac{m}{n}$ is an integer, $n+\frac{m^{2}}{n}$ is also an integer, thus $n=1$. Similarly, $m=1$. Therefore, $k=4$.