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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?

Some comments (building from my notes and also from the comments and answers below)

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. 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 $7494-7278=6^4$.

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.

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).

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.

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?

Some comments (building from my notes and also from the comments and answers below)

The set of even $n$ (nonnegative, naturally) with $F_{2n}$ the sum of two squares begins $$\{ 0,2,6,12,14,26,38,62,74,86,98,122,134,146,158,182,222,254,$$ $$326,338,366,398,446,614,626,698,722,794,866,1022,1046,\ldots\},$$ using the tables at Mersennus.net. 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 $$\ldots, 1082,1226,1238,1418,1646,1814,2174,2246,2258,2282,2294,$$ $$2426,2498,2558,3002,3062,3302,3494,3662,3698,3782,3902,4058,$$ $$4106,4178,4274,4394,4478,4502,4574,4622,4682,4826,4874,4898,4934,$$ $$4946,5102,5174,5558,5594,5702,5714,5798,6074,6326,6362,6542,6614,$$ $$6638,6746,6794,6914,6998,7022,7154,7278,7286,7382,7394,7454,7494,$$ $$7538,7586,7694,7754,7838,7934,8006,8054,8138,8186,8222,8258,8486,$$ $$8522,8594,8906,9038,9074,9194,9206,9242,9326,9398,9446,9638,9662,$$ $$9782,9806,9818,9866,9902$$

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 $7494-7278=6^4$. Also, all four of 222, 366, 7278, 7494 have the form $6p$ (with $p$ a prime, of course).

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.

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).

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.

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?

Some comments (building from my notes and also from the comments and answers below)

The set of even $n$ (nonnegative, naturally) with $F_{2n}$ the sum of two squares begins $$\{ 0,2,6,12,14,26,38,62,74,86,98,122,134,146,158,182,222,254,326,338,366,398,446,\cdots\},$$ using the tables at Mersennus.net. 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 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 78, that are not the sum of two squares but do not seem to be eliminatable in this manner.

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.

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?

Some comments (building from my notes and also from the comments and answers below)

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. 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 $7494-7278=6^4$.

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.

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).

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.

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?

Some comments (building from my notes and also from the comments and answers below)

The set of even $n$ (nonnegative, naturally) with $F_{2n}$ the sum of two squares begins $$\{ 0,2,6,12,14,26,38,62,74,86,98,122,134,146,158,182,222,254,326,338,366,398,446,\cdots\},$$ using the tables at Mersennus.net. 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 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 78, that are not the sum of two squares but do not seem to be eliminatable in this manner.

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?

Some comments (building from my notes and also from the comments and answers below)

The set of even $n$ (nonnegative, naturally) with $F_{2n}$ the sum of two squares begins $$\{ 0,2,6,12,14,26,38,62,74,86,98,122,134,146,158,182,222,254,326,338,366,398,446,\cdots\},$$ using the tables at Mersennus.net. 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 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 78, that are not the sum of two squares but do not seem to be eliminatable in this manner.

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.