<s>[INST] Knowing that $m$ and $n$ are positive integers satisfying $mn \mid m^2 + n^2 + m$, prove that $m$ is a square number. 
Knowing that $m$ and $n$ are positive integers satisfying $$\large mn \mid m^2 + n^2 + m$$, prove that $m$ is a square number.

We have that $mn \mid m^2 + n^2 + m \implies mn \mid (m^2 + n^2 + m)(n + 1)$
$\implies mn \mid m^2n + n^3 + mn + m^2 + n^2 + m$ $\implies \left\{ \begin{align} &mn \mid n^3\\ &mn \mid n^3 + m^2 + n^2 + n \end{align} \right.$
$\implies \left\{ \begin{align} m &\mid n^2\\ mn &\mid m(m + 1) + n^2(n + 1) \end{align} \right.$ $\implies \left\{ \begin{align} mm' &= n^2\\ mn &\mid m(m + 1) + mm'(n + 1) \end{align} \right. (m \in \mathbb Z^+)$
$\implies \left\{ \begin{align} mm' &= n^2\\ m'n &\mid m'(m + 1) + m'^2(n + 1) \end{align} \right.$ $\implies \left\{ \begin{align} mm' &= n^2\\ m'n &\mid n^2 + m' + m'^2 \end{align} \right.$
That also means that if the $m$ is proven to be a square number, $m'$ is also a perfect square. Moreover, $mm' = n^2 \implies (m, m') = 1$ needs to be proven. But I don't know how to.
 [/INST] Note $\ kmn = m^2+m+n^2\,\Rightarrow\, x^2+(1\!-\!kn)\,x + n^2\,$ has roots $\,x = m,m'\in \Bbb Z\,$ satisfying  $ \underbrace{mm' = \color{#0a0}{n^2}}_{\textstyle\small p\mid m\,\Rightarrow\, p\mid\color{#c00} n}\!$ and $\: \underbrace{m+m' = kn-1}_{\textstyle\small p\mid m,m', \color{#c00}n\,\Rightarrow\,p\mid 1 }\,$ thus $\,\color{}{m,m'}\,$ are coprime, hence both are also $\color{#0a0}{\text{squares}}$.
</s>