# Expressing Fibonacci numbers as the sum of squares.

As we know that by observation, the Fibonacci numbers ($F_0=0$, $F_1=1$, $F_{n}=F_{n-1}+F_{n-2}$) have the identity $$F_{2k+1}=F_k^2 + F_{k+1}^2.$$ In particular, if $n$ is odd, then $F_n$ is a sum of two squares. Are there infinitely many even $n$ for which $F_n$ is a sum of two squares? If yes, how one can generalize by giving a proper proof or method?

-
This is only an incomplete (negative) answer. Modulo 4, the the starting 6 terms are (0, 1, 1, 2, 3, 1), and the sequence repeats by cycling over these six numbers again and again. So, $F_{6k+4}$ is $3\mod{4}$, and hence is not a sum of two squares. – Srivatsan Sep 13 '11 at 5:32
Yes. I want to know all such moduli. could you explain... – pavani.neta Sep 13 '11 at 5:55