Find all positive integers such that $n\phi(m)=m\phi(n)$ where $\phi$ is the Euler totient function.
I think that $n\phi(m)=m\phi(n)$ implies $n=m$ and that I should use prime factorization of n and m to show the $n=m$. So after writing n and m in their prime factorization I have: $n\phi(m)=m\phi(n) \Longleftrightarrow nm(\prod \limits_{i=1}^{k}(1-1/p_i)=mn(\prod\limits_{i=1}^{l}(1-1/q_i) \Longleftrightarrow\prod \limits_{i=1}^{k}(1-1/p_i) =\prod\limits_{i=1}^{l}(1-1/q_i)$
Consider the equality $$\prod \limits_{i=1}^{k}\frac{p_i-1}{p_i} =\prod\limits_{i=1}^{l}\frac{q_i-1}{q_i}$$ with $p_1<\ldots <p_k$. Since $p_k$ is in the denominator of the LHS, it must equal to one of the $q_i$. Divide on both sides by $\frac{p_k-1}{p_k}$ and repeat. We get that $k=l$ and $p_i$, $q_i$ are equal in some order. Therefore $\frac{\phi(m)}{m}=\frac{\phi(n)}{n}$ if and only if $m$, $n$ have the same factors