Why is $F^2_{k+2}-F_kF_{k+4}=(-1)^k$ 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.
 A: 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.
A: 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.
A: 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.
