10. [40] Find all pairs $(n, k)$ of positive integers such that
$$
\sigma(n) \phi(n)=\frac{n^{2}}{k}
$$
Answer: $(\mathbf{1}, \mathbf{1})$.
Solution. It is clear that for a given integer $n$, there is at most one integer $k$ for which the equation holds. For $n=1$ this is $k=1$. But, for $n>1$, problem 1 asserts that $\sigma(n) \phi(n) \leq n^{2}-1\frac{n^{2}}{\sigma(n) \phi(n)}$. Write $n=p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$, where the $p_{i}$ are distinct primes and $e_{i} \geq 1$ for all $i$, and let $q_{1}<q_{2}<\cdots$ be the primes in ascending order. Then
$$
\begin{array}{l}
\frac{n^{2}}{\sigma(n) \phi(n)}=\prod_{i=1}^{k} \frac{p_{i}^{2 e_{i}}}{\frac{p_{i}^{e_{i}+1}-1}{p_{i}-1} \cdot\left(p_{i}-1\right) p_{i}^{e_{i}-1}}=\prod_{i=1}^{k} \frac{p_{i}^{2 e_{i}}}{p_{i}^{2 e_{i}}-p_{i}^{e_{i}-1}} \\
=\prod_{i=1}^{k} \frac{1}{1-p_{i}^{-1-e_{i}}} \leq \prod_{i=1}^{k} \frac{1}{1-p_{i}^{-2}}<\prod_{i=1}^{\infty} \frac{1}{1-q_{i}^{-2}} \\
=\prod_{i=1}^{\infty}\left(\sum_{j=0}^{\infty} \frac{1}{q_{i}^{2 j}}\right)=\sum_{n=1}^{\infty} \frac{1}{n^{2}} \\
<1+\sum_{n=2}^{\infty} \frac{1}{n^{2}-1}=1+\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots\right)=\frac{7}{4}<2
\end{array}
$$

It follows that there can be no solutions to $k=\frac{n^{2}}{\sigma(n) \phi(n)}$ other than $n=k=1$.