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I was doing my math exercise and found out a problem.

It is

Is there infinitely many pairs of integers $(m,n)$ such that $m|n^2+1$ and $n|m^2+1$?

After some trials and tips from friends, I found that $(m.n)=(F_{2k-1}, F_{2k+1})$ is a solution. And the friend told me to prove that $F^2_{k+2}-F_kF_{k+4}=(-1)^k$ to finish the solution. But I cannot find any method to prove that (maybe I am dumber than you guys). Can someone help me?

Note that $F_k$ is the $k$th Fibonacci-number.

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  • $\begingroup$ What does $F_k$ stands for ? $\endgroup$ – Fallen_Prince Nov 16 '19 at 11:55
  • $\begingroup$ These are the Fibonacci numbers! $\endgroup$ – Dr. Sonnhard Graubner Nov 16 '19 at 11:57
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    $\begingroup$ The sledgehammer is to use Binet's formula. $\endgroup$ – Angina Seng Nov 16 '19 at 12:01
  • $\begingroup$ Hey, @Culver Kwan don't ask the question asked by me!!!! Also, you have not done any trials ok? $\endgroup$ – Isaac YIU Math Studio Nov 16 '19 at 12:51
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This is a special case of Catalan's identity. See the proofs here https://proofwiki.org/wiki/Catalan%27s_Identity

Replacing $n$ with $k+2$ and plugging in $r=2$ into any of them and finally noting that $F_2=1$ you get the proof of your claim.

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If we shift the indices in the question by $2$ and rearrange, an equivalent identity is called "Cassini's Identity": $$(-1)^k = F_{k+1} F_{k-1} - F_k^2$$

One way to prove Cassini's Identity is to start with a matrix version of the definition of the Fibonacci sequence: $$\begin{pmatrix} 1 &1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix} F_{k+1} &F_{k} \\ F_{k} &F_{k-1} \end{pmatrix}$$ for $k \ge 1$, which you can easily prove by induction on $k$. Then taking determinants of both sides of the equation yields Cassini's Identity.

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A simple direct proof

Start with any pair of positive integers $m<n$ such that $m|n^2+1$ and $n|m^2+1$. All we need prove is that from any such pair we can construct a larger pair.

Let $m^2+1=kn$ and $n^2+1=lm,$ then $n<l$.

$l^2+1=(\frac{n^2+1}{m})^2+1=\frac{n^4+2n^2+1+m^2}{m^2}=\frac{n(n^3+2n+1+k)}{m^2}$

$m$ and $n$ are coprime and therefore $n|l^2+1$. Thus $l$ and $n$ are the larger solutions we require.

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