Divisibility of unknown numbers

$\left\{\begin{matrix}m^2+n \vdots n^2-m\\ m+n^2 \vdots m^2-n\end{matrix}\right.$ $(1)$

$\Leftrightarrow \left\{\begin{matrix} m^2+n \geq n^2-m\\ m+n^2 \geq m^2-n \end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix} (m-n+1)(m+n) \geq 0\\ (n-m+1)(m+n) \geq 0 \end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix} m-n+1 \geq 0\\ n-m+1\geq 0 \end{matrix}\right.$ (Because $m,n$ are positive integers, $m+n>0$ )

$\Leftrightarrow -1 \leq m-n \leq 1$

As $m,n$ are positive integers, $m-n \in \{ -1,0,1\}$

$\text{Case 1: } m-n=-1 \Leftrightarrow m=n-1$

$n^2+m \vdots m^2-n \Rightarrow \frac{n^2+m}{m^2-n} \in \Z$

$\Leftrightarrow \frac{n^2+n-1}{(n-1)^2-n} \in \Z$

$\Leftrightarrow \frac{n^2+n-1}{n^2-3n+1} \in \Z$

$\Leftrightarrow \frac{(n^2-3n+1) + 4n - 2}{n^2-3n+1} \in \Z \Rightarrow \frac{4n-2}{n^2-3n+1} \in \Z$

$\Rightarrow 4n-2 \geq n^2-3n+1 \Leftrightarrow n^2-7n+3 \leq 0$

$\Rightarrow \frac{7-\sqrt{37}}{2} \leq n \leq \frac{7+\sqrt{37}}{2}$

As $n$ is a positive integer, $n \in \{ 1,2,3,4,5,6\}$

Because $m=n-1$ , we have $(m,n) = (1,2),(2,3),(3,4),(4,5),(5,6)$

Substitute these results into $(1)$ , the solutions satisfying $(1)$ are $(m,n) = (1,2) , (2,3)$

$\text{Case 2: } m-n=0 \Leftrightarrow m=n$

$\frac{n^2+m}{m^2-n} \in \Z$

$\Leftrightarrow \frac{n^2+n}{n^2-n} \in \Z$

$\Leftrightarrow \frac{(n^2-n) + 2n}{n^2-n} \in \Z \Rightarrow \frac{2n}{n(n-1)} \in \Z \Leftrightarrow \frac{2}{n-1} \in \Z$

$\Rightarrow n-1 \leq 2 \Rightarrow n \leq 3$

As $n$ is a positive integer, $n \in \{ 1,2,3\}$

Because $m=n$ , we have $(m,n) = (1,1),(2,2),(3,3)$

Substitute these results into $(1)$ , the solutions satisfying $(1)$ are $(m,n) = (2,2), (3,3)$

$\text{Case 3: } m-n=1 \Leftrightarrow m=n+1$

$m^2+n \vdots n^2-m \Rightarrow \frac{m^2+n}{n^2-m} \in \Z$

$\Leftrightarrow \frac{(n+1)^2+n}{n^2-n-1} \in \Z$

$\Leftrightarrow \frac{n^2+3n+1}{n^2-n-1} \in \Z$

$\Leftrightarrow \frac{(n^2-n-1) + 4n + 2}{n^2-n-1} \in \Z \Rightarrow \frac{4n+2}{n^2-n-1} \in \Z$

$\Rightarrow 4n+2 \geq n^2-n-1 \Leftrightarrow n^2-5n-3 \leq 0$

$\Rightarrow \frac{5-\sqrt{37}}{2} \leq n \leq \frac{5+\sqrt{37}}{2}$

As $n$ is a positive integer, $n \in \{ 1,2,3,4,5\}$

Because $m=n+1$ , we have $(m,n) = (2,1),(3,2),(4,3),(5,4),(6,5)$

Substitute these results into $(1)$ , the solutions satisfying $(1)$ are $(m,n) = (2,1) , (3,2)$

$\text{Conclusion: }$ There are $\boxed{6}$ solutions satisfying $(1)$ , which are $(m,n) = (1,2), (2,1), (2,2), (2,3), (3,2), (3,3)$

Code

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for n in range(1,180):
    for m in range(1,180):
        if (m**2!=n):
            if (n**2!=m):
                if ((m**2+n)%(n**2-m)==0):
                    if ((n**2+m)%(m**2-n)==0):
                        print (m,n)

Output

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2 1
1 2
2 2
3 2
2 3
3 3