# 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.

• What does $F_k$ stands for ? – Fallen_Prince Nov 16 '19 at 11:55
• These are the Fibonacci numbers! – Dr. Sonnhard Graubner Nov 16 '19 at 11:57
• The sledgehammer is to use Binet's formula. – Angina Seng Nov 16 '19 at 12:01
• Hey, @Culver Kwan don't ask the question asked by me!!!! Also, you have not done any trials ok? – Isaac YIU Math Studio Nov 16 '19 at 12:51

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.

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 simple direct proof

Start with any pair of positive integers $$m such that $$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.

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

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