Prove there exists $m$ and $k$ such that $ n = mk^2$ where $m$ is not a multiple of the square of any prime. For any positive integer $n$, prove that there exists integers $m$ and $k$ such that:
$$n = mk^2 $$
where $m$ is not a multiple of the square of any prime. (For all primes $p$, $p^2$ does not divide $m$)
I didn't have much success proving this inductively, any suggestion?

The prime decomposition of $n = \prod p_i^{a_i}$. Set $m= \prod_{odd\: a_i} p_i$; then $\frac{n}{m}$ has only even-exponent prime factors (if any) and can be represented as $k^2$ (which may possibly be equal to $1$).
