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?

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Some comments (building from my notes and also from the comments and answers below)
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The set of even $n$ (nonnegative, naturally) with $F_{2n}$ the sum of two squares begins
  $$\{ 0,2,6,12,14,26,38,62,74,86,98,122,134,146,158,182,222,254,$$
  $$326,338,366,398,446,614,626,698,722,794,866,1022,1046,\cdots\},$$
using the tables at [Mersennus.net][1]. There are five entries in this list that are not 2 mod 12: three small numbers (0, 6, 12) that can be forgiven their impertinence, but also 222 and 366 (both are 6 mod 12, and also 78 mod 144).

The list of possible indices (possible because not all of the factorizations are complete) continues
  $$\cdots 1082,1226,1238,1418,1646,1814,2174,2246,2258,2282,2294,$$
  $$2426,2498,2558,3002,3062,3302,3494,3662,3698,3782,3902,4058,$$
  $$4106,4178,4274,4394,4478,4502,4574,4622,4682,4826,4874,4898,4934,$$
  $$4946,5102,5174,5558,5594,5702,5714,5798,6074,6326,6362,6542,6614,$$
  $$6638,6746,6794,6914,6998,7022,7154,7278,7286,7382,7394,7454,7494,$$
  $$7538,7586,7694,7754,7838,7934,8006,8054,8138,8186,8222,8258,8486,$$
  $$8522,8594,8906,9038,9074,9194,9206,9242,9326,9398,9446,9638,9662,$$
  $$9782,9806,9818,9866,9902$$

This list contains two more indices that are not 2 mod 12: 7278 (possibly giving a sum of two squares) and 7494 (definitely giving a sum of two squares). Note $366-222=12^2$ and $7494-7278=6^4$.

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The Fibonacci numbers are periodic modulo $m$ (for any $m>1$). Considering the sequence modulo 4, for example, it repeats 0, 1, 1, 2, 3, 1. Since the sum of two squares is never 3 mod 4, we learn that $F_{6n+4}$ is never the sum of two squares. Varying the modulus allows us to eliminate many other congruence classes. There are some numbers, for example $F_{78}, that are not the sum of two squares but do not seem to be eliminatable in this manner.

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If $n$ is negative and even, then $F_n$ is negative. This restricts the possibilities for an algebraic family (such as the one that exists for odd-indices).

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The Lucas numbers are $L_0=2,L_1=1,L_{n}=L_{n-1}+L_{n-2}$. The identity $F_{2n}=F_nL_n$, coupled with the easy fact that $\gcd(F_n,L_n)$ is 1 or 2, implies that $F_{2n}$ is the sum of two squares if and only if both $F_n$ and $L_n$ are. 

  [1]: http://mersennus.net/fibonacci/