# 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

## 2 Answers

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Also see the question there and note the similarity. –  Dan Brumleve Sep 13 '11 at 5:37
Yes, I have seen now. I want more clarety. –  pavani.neta Sep 13 '11 at 5:52
pavani.neta, then you should have explained what you would like to know in the question, instead of copying it from another user on MO without attribution. Also consider leaving a comment on one of the answers over there. –  Dan Brumleve Sep 13 '11 at 5:59
@pav The second and third sentences above are exactly the same as the MO question. Ditto for "have the identity" [sic] in the first sentence. –  Bill Dubuque Sep 13 '11 at 6:21
"Make it more clear" - what does that mean? There is a lot of stuff there at MO - can you be precise about what is unclear to you? And if you "can prove the same by better method", why does it matter that it's unclear? Why not just post your better method here, or to MO? –  Gerry Myerson Sep 13 '11 at 7:35

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!

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