I am no mathematician so my apologies for my ignorance.
I notice that every number in the Fibonacci series can be expressed as a previous Fibonacci number squared plus or minus (alternating) another previous Fibonacci number squared, and both columns of squares themselves are composed of the Fibonacci sequence, each number appearing twice from 2 onwards. ie
1 1 + 0 1+0
1 1 - 0 1-0
2 1 + 1 1+1
3 2 - 1 4-1
5 2 + 1 4+1
8 3 - 1 9-1
13 3 + 2 9+4
21 5 - 2 25-4
34 5 + 3 25+9
55 8 - 3 64-9
89 8 + 5 64+25
144 13 - 5 169-25
233 13 + 8 169+64
377 21 - 8 441-64
610 21 + 13 441+169
987 34 - 13 1156-169
and so on.
Is this well known? Is there a proof that it works for all Fibonacci numbers?
It a derivation from the Catalan and Cassini identity. Cassini identity states that:
$$F_n^2 - F_{n-r} \cdot F_{n+r} = (-1)^{n-r}F_r^2$$
Now set $r=n-1$ and we got:
$$F_n^2 - (-1)^{n-n+1}F_{n-1}^2 = F_{n - n+1}F_{n+n-1}$$ $$F_n^2 + F_{n-1}^2 = F_{2n - 1}$$
But $2n - 1$ is odd, so this covers the case when the Fibonacci index is odd number. Now by setting $r=n-2$ we have:
$$F_n^2 - (-1)^{n-n+2}F_{n-2}^2 = F_{n - n+2}F_{n+n-2}$$ $$F_n^2 - F_{n-2}^2 = F_{2n - 2}$$
This one covers the case when the Fibonacci index is even.