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#1 h Apr 16, 2025, 1:53 PM

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Find all $m$ and $n$ such that $m n + 1 | m^2 - n^2,$ where $m, n$ are integers and $m > n$ .

This post has been edited 2 times. Last edited by benjaminchew13, Apr 25, 2025, 7:22 AM

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#2 h Apr 16, 2025, 1:54 PM

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Well the solutions that i have found are $m = n^2 + n + 1$ , and any $m, \text{but }n = 1$

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#3 h Apr 17, 2025, 1:29 PM

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This is equivalent to find which factors of $n^4-1$ have the form $mn+1$ , which I think to be an open problem.

Indeed, we can write it as $0 = m^2 - n^2 = m^2 - n^2 (mn)^2 = m^2 (1-n^4) \text{(mod } mn+1 \text{)},$ which gives $\frac{n^4-1}{mn+1}$ is an integer.

Some solutions outside of your set are $(5,3)$ , $(17,7)$ , which are difficult to formulae.

A similar problem (but solvable) is IMO shortlist 1994 N2.

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#4 h Apr 20, 2025, 12:54 PM

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Using Vieta jumping we can find lots of solutions. I will look at integer solutions, not only positive ones. So suppose ${{m^2-n^2}\over {mn+1}}=k$ .
We can do $(m,n)\rightarrow (n,m)$ and $(m,n)\rightarrow (-m,-n)$ but there is more.
Rewrite as $m^2-kmn-(n^2+k)=0$ then the two solutions $m$ for the quadratic equation satisfy $m_1+m_2=kn$ .
Hence we can also do $(m,n)\rightarrow (kn-m,n)$ to find another solution for the same $k$ .
Looking at $n$ instead of $m$ we also find $(m,n)\rightarrow (m,-km-n)$ .
Now we can also do compositions: $(m,n)\rightarrow (kn-m,n)\rightarrow (kn-m, km-k^2n-n)$ .

For example. start with $(1,n)$ and $k=1-n$ . Then $(m,n)\rightarrow (n-n^2-1,-n^3+2n^2-3n+1)$ and flipping signs we find $(n^2-n+1,n^3-2n^2+3n-1)$ . Then $n=2,3$ yield the solutions $(5,3),(17,7)$ which were mentioned in #3. I'll have a look at the problem later, perhaps all solutions can be found with this procedure. The problem also ressembles an old IMO problem from 1988, which considers the solutions to ${{a^2+b^2}\over {ab+1}}\in Z$ .

This post has been edited 3 times. Last edited by alexheinis, Apr 20, 2025, 1:03 PM

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