How to Prove This Fibonacci Identity Involving Squares and Products?

[ramsey2879]

Can someone guide me on how to prove that
$$F_{4n+3} + F_{4n+6} = F_{2n+1}^2 + F_{2n+4}^2$$

either side of the above is the difference

$$(F_{2n+2}*F_{2n+3} + F_{2n+4}^2) - (F_{2n}*F_{2n+1} + F_{2n+2}^2)$$

I intend to post this sequence $$F_{2n}*F_{2n+1} + F_{2n+2}^2$$, with a comment re a few properties thereof, on Sloane's online encyclopedia of integer sequences but would like to verify the above identity first.

 

[tehno]

First thing springs to mind is to try use general formula for n^th
Fibbonacci number:

$$F_{n}=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$$

In order to proceed with math induction.But I'm unsure will it work or not.
I'm sure there are better methods ,though.

 

[ramsey2879]

First thing springs to mind is to try use general formula for n^th
Fibbonacci number:

$$F_{n}=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$$

In order to proceed with math induction.But I'm unsure will it work or not.
I'm sure there are better methods ,though.

Thanks
I think there is an identity for the following that works:

$$F_{i}*F_{j} + F_{i+1}*F_{j+1} = F_{?}$$

Let j = i = 2n+1 then

$$F_{2n+1}^{2} + F_{2n+2}^{2} = F_{4n+3}$$
$$F_{2n+2}^{2} + F_{2n+3}^{2} = F_{4n+5}$$
$$F_{2n+2}^{3} + F_{2n+4)^{2} = F_{4n+7}$$
\\
$$F_{4n+3} +F_{4n+6} = F_{4n+3} + F_{4n+7} - F_{4n+5}$$
$$=F_{2n+1}^{2} + F_{2n+4}^{2}$$