# 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?

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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
F[6] = 8 = 4 + 4 = F[3]^2 + F[3]^2 –  Random Excess May 16 at 23:27

Actually, what applies to $F_{2k+1}$ does not apply to $F_{2k}$ at all. Here is the identity for $F_{2k}$: \begin{align} F_{2k}&=F_kF_{k-1}+F_kF_{k+1}\\ &=F_k(F_{k-1}+F_{k+1})\\ &=F_kL_k \end{align} Therefore, there are no $F_n$ with even $n$ where $F_n$ is the sum of two squares of Fibonacci numbers. I don't know if this means that $F_n$ is strictly never a sum of two squares as long as $n$ is not odd, but certainly not squares of Fibonacci numbers. But it is also true that \begin{align}F_{2k}&=F_{2k+1}-F_{2k-1}\\ &=F_{k+1}^2+F_k^2-F_k^2-F_{k-1}^2\\ &=F_{k+1}^2-F_{k-1}^2\end{align} So there are infinitely many even $n$ for which $F_n$ is a difference of two squares!