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Suppose $m$ and $n$ are integers (without loss of generality, nonnegative integers) satisfying the relations $m|n^2-1$ and $n|m^2-1$. What can be said about $m$ and $n$?

One family of solutions can be constructed as follows: let $k$ be an arbitrary integer, and consider the sequence $a^k_n$ defined recursively by $$ a^k_0=0\\ a^k_1=1\\ a^k_j=ka^k_{j-1}-a^k_{j-2} $$

This sequence satisfies the "Cassini-like" identity $$ a^k_{j+1}a^k_{j-1}=(a^k_j)^2-1 $$ which means that we can take $(m,n)=(a^k_j,a^k_{j+1})$ for any $k,j$.

Note that this family includes the obvious solutions $n=m+1$, which are consecutive values of the sequence $a^2_j$. It also includes the slightly-less-obvious solutions $n=m^2-1$, as $a^k_2=k$ and $a^k_3=k^2-1$.

A computer search shows that every pair $(m,n)$ satisfying these relations with $m,n<10^5$ is of this form.

Is every pair which satisfies these relations of this form?

(This question arose out of an early attempt to solve this problem, before I found some stronger divisibility conditions which led to a complete solution.)

Micah
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  • Related: https://math.stackexchange.com/questions/1917044/proving-the-converse-of-an-imo-problem-are-there-infinitely-many-pairs-of-posit?rq=1 – Micah Apr 30 '17 at 17:48

1 Answers1

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Given $m,n$ with $m<n$, write $p=\frac{m^2-1}{n}$. We clearly have $p<m$.

Now, we se that $np\equiv -1\pmod m$. Since $n^2\equiv 1\pmod m$ this means that $p^2\equiv 1\pmod{m}$. So $m\mid p^2-1$.

But we also have that $p\mid m^2-1$. So we have a "smaller" solution, $(p,m)$.

This lets us see that the sequence must be of the form:

$$a_0=0,a_1=1, a_2=?, a_{n+1}=\frac{a_n^2-1}{a_{n-1}}$$

The value of $a_2$ then determines $k$ - that is, $a_2=k$.

Then it is just an induction proof to show that this $a_i$ is your $a_i^k$.

Then consider the case $m=n$.

Thomas Andrews
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