For $F(n)$ the $n$-th Fibonacci number, is $F(a)F(b)-F(a+1)F(b-1)$ always $\pm F(m)$ for some $m$?

Question

For $F(n)$ the $n$th Fibonacci number, the expression $$F(a)F(b)-F(a+1)F(b-1)$$

seems to be $\pm F(m)$ for some $m$. I can't specify $m$ or the sign in terms of $a,b,$ and have not tried it out extensively. My question is:

Is there a link to such a formula (and/or its proof) that someone could cite? I'd appreciate it.

I've tried via Binet formulas, and several terms cancelled, but I couldn't get it to go through. Thanks for any help.

Answer 1

We have d'Ocagne's identity $$F(a)F(b)-F(a+1)F(b-1)= (-1)^{b-1}F(a-b+1)$$

Googleing: http://nntdm.net/papers/nntdm-20/NNTDM-20-5-44-48.pdf

Answer 2

For your information, a similar identity can be found in Fibonacci and Lucas Numbers, and the Golden Section by Steven Vajda, Dover, 1989.

$$F_{n+h}F_{n+k}-F_{n}F_{n+h+k}=(-1)^{n}F_{h}F_{k}$$