Let $f : N \to N$ such that $f(p) = 1$ for all p prime and $f(ab) =bf(a) + af(b)$ for all $a, b \in N$. Prove that if $n = p^{a_1}_1 p^{a_1}_2... p^{a_1}_k$ is the canonical distribution of $n$ and $p_i$ does not divide $a_i$ ($i = 1, 2, ..., k$) then $\frac{n}{gcd(n,f(n))}$ is square free (not divisible by a square greater than $1$).