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Claim: The density of Fibonacci numbers which are not divisible by a prime of the form $3+4k$ is $\frac{1}{2}$.

Proof: The odd ones are not, so this gives at least $\frac{1}{2}$. Since the density of numbers which are not divisible by a prime of the form $5+6k$ is zero, it follows from the previous claim that the density of even Fibonacci numbers not divisible by a prime of the form $3+4k$ is $0$.

$F_{12}$ is a very special Fibonacci number for a few reasons. One is that it is the only nontrivial square. If we change the condition to a sum of two nonzero squares, then $F_{12}$ is automatically excluded, so we can rephrase conjecture 1 as:

Claim: The density of even Fibonacci numbers which are not divisible by some prime of the form $3+4k$ is $0$.

Proof: This is a corollary of Will Jagy's observation. Since the density of numbers which are not divisible by a prime of the form $5+6k$ is zero, it follows from the previous claim that the density of even Fibonacci numbers not divisible by a prime of the form $3+4k$ is $0$.

$F_{12}$ is a very special Fibonacci number for a few reasons. One is that it is the only nontrivial square. If we change the condition to a sum of two nonzero squares, then $F_{12}$ is automatically excluded. Also, since no other Fibonacci numbers are squares, nothing else is affected. Hence we can rephrase conjecture 1 as:

This is not a solution, just some thoughts which are too long for a comment. I think they could possibly be useful for someone, so I am undeleting this answer. (Tell me if you think I should reconsider)

The Fibonacci numbers are a divisibility sequence, which means that $$\gcd(F_n,F_m)=F_{\gcd(n,m)}.$$ Also recall that we can write $F_n$ as a sum of two squares if it has no prime factors of the form $3+4k$.

Previously, I said some things about the above being if and only if for primes of the form $3+4k$ (it clearly isn't for $1+4k$) Small update: It also is just false for $3+4k$, since $3571=3+4k$, is prime and divides $F_{68}$.

Added: Proof of part of Will Jagy's Observation:

Claim: If $n\equiv 5\pmod{6}$ then $L_n$, and hence $F_{2n}$ are divisible by some prime $p\equiv 3 \pmod{4}$.

Proof: Look at $L_n$ modulo $4$. Then the sequence is $L(0)\equiv 2$, $L(1)\equiv 1$, $L(2)\equiv 3$, $L(3)\equiv 0$, $L(4)\equiv 3$, $L(5)\equiv 3$, $L(6)\equiv 2$, $L(7)\equiv 1$, and at this point it must repeat. The cycle length is $6$, and $L(5)\equiv 3$. This means that $L(5+6k)\equiv 3\pmod{4}$ for all $k$. Hence $L(5+6k)$ is always divisible by a prime congruent to $3$ mod $4$.

Since $L_n |L_{kn}$ when $k$ is odd, we can conclude that if $p\equiv 5 \pmod{6}$ divides $n$, then some prime $q\equiv 3\pmod{4}$ must divide $F_{2n}$. This is because either $2|n$, and hence $3|F_{2n}$, or $n$ is odd, and $L_p|L_n$ so that $q|F_{2n}$.

Example: The first few primes congruent to $3$ mod $4$ less then $100$ are $$3,7,11,19,23,31,41,47,59,67,71,79,83,87$$ and they divide respectively $$F_4, F_8, F_{10}, F_{18}, F_{24}, F_{30}, F_{40}, F_{48}, F_{58}, F_{68}, F_{70}, F_{78}, F_{84}, F_{88}.$$

This is not a solution, just some thoughts which are too long for a comment. I have added proofs of Will Jagy and Junkies comments/conjectures which are fairly interesting on their own.

First, the Fibonacci numbers are a divisibility sequence, which means that $$\gcd(F_n,F_m)=F_{\gcd(n,m)}.$$
Added: Proof of part of Will Jagy's Observation:

Claim: If $n\equiv 5\pmod{6}$ then $L_n$, and hence $F_{2n}$ are divisible by some prime $p\equiv 3 \pmod{4}$.

Proof: Look at $L_n$ modulo $4$. Then the sequence is $L(0)\equiv 2$, $L(1)\equiv 1$, $L(2)\equiv 3$, $L(3)\equiv 0$, $L(4)\equiv 3$, $L(5)\equiv 3$, $L(6)\equiv 2$, $L(7)\equiv 1$, and at this point it must repeat. The cycle length is $6$, and $L(5)\equiv 3$. This means that $L(5+6k)\equiv 3\pmod{4}$ for all $k$. Hence $L(5+6k)$ is always divisible by a prime congruent to $3$ mod $4$.

Since $L_n |L_{kn}$ when $k$ is odd, we can conclude that if $p\equiv 5 \pmod{6}$ divides $n$, then some prime $q\equiv 3\pmod{4}$ must divide $F_{2n}$. This is because either $2|n$, and hence $3|F_{2n}$, or $n$ is odd, and $L_p|L_n$ so that $q|F_{2n}$.

Added Proof Of Junkie's Comment:

Claim: The density of Fibonacci numbers which are not divisible by a prime of the form $3+4k$ is $\frac{1}{2}$.

Proof: The odd ones are not, so this gives at least $\frac{1}{2}$. Since the density of numbers which are not divisible by a prime of the form $5+6k$ is zero, it follows from the previous claim that the density of even Fibonacci numbers not divisible by a prime of the form $3+4k$ is $0$.

Conjectures and other thoughts: Recall that we can write $F_n$ as a sum of two squares if it has no prime factors of the form $3+4k$.

Previously, I said some things about what happens if the above were an "if and only if" for primes of the form $3+4k$ (it clearly isn't for $1+4k$) Small update: It also is just false for $3+4k$, since $3571=3+4k$, is prime and divides $F_{68}$.

Examples: The first few primes congruent to $3$ mod $4$ less then $100$ are $$3,7,11,19,23,31,41,47,59,67,71,79,83,87$$ and they divide respectively $$F_4, F_8, F_{10}, F_{18}, F_{24}, F_{30}, F_{40}, F_{48}, F_{58}, F_{68}, F_{70}, F_{78}, F_{84}, F_{88}.$$

This is not a solution, just some thoughts which are too long for a comment. I think they could possibly be useful for someone, so I am undeleting this answer. (Tell me if you think I should reconsider)

The Fibonacci numbers are a divisibility sequence, which means that $$\gcd(F_n,F_m)=F_{\gcd(n,m)}.$$ Also recall that we can write $F_n$ as a sum of two squares if it has no prime factors of the form $3+4k$.

Conjecture 1: The only Fibonacci number of the form $F_{2n}$ which is divisible by some prime of the form $3+4k$ and can be written as the sum of two squares is $F_{12}$.

$F_{12}$ is a very special Fibonacci number for a few reasons. One is that it is the only nontrivial square. If we change the condition to a sum of two nonzero squares, then $F_{12}$ is automatically excluded, so we can rephrase conjecture 1 as:

Conjecture 1: If $F_{2n}$ is divisible by some prime of the form $3+4k$ then it cannot be written as the sum of two nonzero squares.

One reason for this conjecture is that it checks out numerically in large range, up to $F_{1000}$. Also, it would be nice if it was true.

Assuming this, by the divisibility property, given a prime $p=3+4k$, we need only care about the first time it appears in the Fibonacci Sequence. There is a theorem which states that $$F_{p-\left(\frac{p}{5}\right)}\equiv 0 \pmod{p},\ \ \ \ \ \ \ \ \ (1) $$ where $\left(\frac{p}{5}\right)$ is the Legendre symbol. Conjecture 1 would imply that if $2n$ is divisible by $p-\left(\frac{p}{5}\right)$ for any $p\equiv 3 \pmod{4}$, then $F_{2n}$ is not the sum of two nonzero squares.

Previously, I said some things about the above being if and only if for primes of the form $3+4k$ (it clearly isn't for $1+4k$) Small update: It also is just false for $3+4k$, since $3571=3+4k$, is prime and divides $F_{68}$.

Example: The first few primes congruent to $3$ mod $4$ less then $100$ are $$3,7,11,19,23,31,41,47,59,67,71,79,83,87$$ and they divide respectively $$F_4, F_8, F_{10}, F_{18}, F_{24}, F_{30}, F_{40}, F_{48}, F_{58}, F_{68}, F_{70}, F_{78}, F_{84}, F_{88}.$$

So in particular, (assuming conjecture 1) if I want $F_{2n}$ to be a sum of two nonzero squares, $2n$ cannot be divisible by any of the above. I.e. we cannot have any of $$2|n, \ 5|n, \ 9|n,\ 29|n,\ 39|n .$$ This idea can give us a large list of primes, none of which can divide $n$, but that is about all I can get it to do.

This is not a solution, just some thoughts which are too long for a comment. I think they could possibly be useful for someone, so I am undeleting this answer. (Tell me if you think I should reconsider)

The Fibonacci numbers are a divisibility sequence, which means that $$\gcd(F_n,F_m)=F_{\gcd(n,m)}.$$ Also recall that we can write $F_n$ as a sum of two squares if it has no prime factors of the form $3+4k$.

Conjecture 1: The only Fibonacci number of the form $F_{2n}$ which is divisible by some prime of the form $3+4k$ and can be written as the sum of two squares is $F_{12}$.

$F_{12}$ is a very special Fibonacci number for a few reasons. One is that it is the only nontrivial square. If we change the condition to a sum of two nonzero squares, then $F_{12}$ is automatically excluded, so we can rephrase conjecture 1 as:

Conjecture 1: If $F_{2n}$ is divisible by some prime of the form $3+4k$ then it cannot be written as the sum of two nonzero squares.

One reason for this conjecture is that it checks out numerically in large range, up to $F_{1000}$. Also, it would be nice if it was true.

Assuming this, by the divisibility property, given a prime $p=3+4k$, we need only care about the first time it appears in the Fibonacci Sequence. There is a theorem which states that $$F_{p-\left(\frac{p}{5}\right)}\equiv 0 \pmod{p},\ \ \ \ \ \ \ \ \ (1) $$ where $\left(\frac{p}{5}\right)$ is the Legendre symbol. Conjecture 1 would imply that if $2n$ is divisible by $p-\left(\frac{p}{5}\right)$ for any $p\equiv 3 \pmod{4}$, then $F_{2n}$ is not the sum of two nonzero squares.

Previously, I said some things about the above being if and only if for primes of the form $3+4k$ (it clearly isn't for $1+4k$) Small update: It also is just false for $3+4k$, since $3571=3+4k$, is prime and divides $F_{68}$.

Added: Proof of part of Will Jagy's Observation:

Claim: If $n\equiv 5\pmod{6}$ then $L_n$, and hence $F_{2n}$ are divisible by some prime $p\equiv 3 \pmod{4}$.

Proof: Look at $L_n$ modulo $4$. Then the sequence is $L(0)\equiv 2$, $L(1)\equiv 1$, $L(2)\equiv 3$, $L(3)\equiv 0$, $L(4)\equiv 3$, $L(5)\equiv 3$, $L(6)\equiv 2$, $L(7)\equiv 1$, and at this point it must repeat. The cycle length is $6$, and $L(5)\equiv 3$. This means that $L(5+6k)\equiv 3\pmod{4}$ for all $k$. Hence $L(5+6k)$ is always divisible by a prime congruent to $3$ mod $4$.

Since $L_n |L_{kn}$ when $k$ is odd, we can conclude that if $p\equiv 5 \pmod{6}$ divides $n$, then some prime $q\equiv 3\pmod{4}$ must divide $F_{2n}$. This is because either $2|n$, and hence $3|F_{2n}$, or $n$ is odd, and $L_p|L_n$ so that $q|F_{2n}$.

Example: The first few primes congruent to $3$ mod $4$ less then $100$ are $$3,7,11,19,23,31,41,47,59,67,71,79,83,87$$ and they divide respectively $$F_4, F_8, F_{10}, F_{18}, F_{24}, F_{30}, F_{40}, F_{48}, F_{58}, F_{68}, F_{70}, F_{78}, F_{84}, F_{88}.$$

So in particular, (assuming conjecture 1) if I want $F_{2n}$ to be a sum of two nonzero squares, $2n$ cannot be divisible by any of the above. I.e. we cannot have any of $$2|n, \ 5|n, \ 9|n,\ 29|n,\ 39|n .$$ This idea can give us a large list of primes, none of which can divide $n$, but that is about all I can get it to do.