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?
 A: Given $n$, there are only finitely many $(k,m)$ such that $n=mk^2$ (one of the solutions being $(1,n)$). Pick a solution $(k,m)$ with maximal $k$. Claim: $m$ is squarefree. Indeed, assume $m=p^2t$. Then the solution $(pk,t)$ contradicts the maximality of $k$.
A: 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 $1$).
A: By induction: $ $ no $\,p^2\mid n\,\Rightarrow\, n = n\cdot 1^2,\ $ else induct $\,\ \begin{align}{n/p_i^2 = mk^2\\ {\rm no}\ \ p^2\mid m}\quad \end{align}\ $ so $\ n = m(p_ik)^2$
Remark $\ $ The proof works not only for the set of squares but for any set $\,S\subset \Bbb N_{>0}\,$ that is closed under $\rm\color{#0a0}{ multiplication}$ and contains $\,\color{#c00}1\,$ (i.e. any monoid). Indeed
By induction: $ $ no $\,s\mid n\,\Rightarrow\, n = n\cdot\color{#c00} 1,\ $ else induct $\,\ \begin{align}{n/s_i = ms_j\\ {\rm no}\ \ s\mid m}\quad \end{align}\ $ so $\ n = m(\color{#0a0}{s_i s_j})$
