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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.

Lê Thành Đạt
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  • How do you prove some number is a square? Your idea of so many implications seem to be confusing you... just convince yourself what you want to conclude to prove m is a square.. what property should that satisfy? – Praphulla Koushik Sep 02 '19 at 17:09
  • Hint: show $m(qn-1-m)=n^2$, where $q=\frac{m^2+n^2+m}{mn}$. – Aphelli Sep 02 '19 at 17:11
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    You have a typo - at one point you wrote $mn \mid n^3+m^2+n^2+n$, but it should say $mn \mid n^3+m^2+n^2+m$. – kccu Sep 02 '19 at 17:13
  • How do you arrive at the $\begin{cases}\end{cases}$? And what is its supposed meaning? "And"? "Or"? – Hagen von Eitzen Sep 02 '19 at 17:18

2 Answers2

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Let $p$ be prime and let $k\ge0 $ be maximal with $p^k\mid m$. Similarly, let $l\ge 0$ be maximal with $p^l\mid n$. So $m=p^kr$, $n=p^ls$ for some positive integers $r,s$. So $$p^{k+l}rs\mid p^{2k}r^2+p^{2l}s^2+p^kr. $$ Conclude that $2l\ge k$ and
$$p^{l}rs\mid p^{k}r^2+p^{2l-k}s^2+r. $$ of $k>0$ then also $l>0$ and so $p\mid p^{2l-k}s+r$, which leads to $2l-k=0$. Therefore $k$ must be even. As this holds for all primes $p$, we conclude that $m$ is a perfect square.

Hagen von Eitzen
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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 $\color{#0a0}{\text{squares}}$.

Bill Dubuque
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