If $F_m = 2^{2^m} + 1$ is prime (with $m \geq 1$), does it follow that $3 \mid \frac{F_m + 1}{2}$? If $$F_m = 2^{2^m} + 1$$ is prime with $m \geq 1$, does it follow that $$3 \mid \frac{F_m + 1}{2}?$$
I have verified this to be true for $1 \leq m \leq 4$.

$F_m=2^{2^m}+1$ are also known as Fermat numbers. From 
$$2 \equiv -1 \pmod{3} \Rightarrow 2^{2^m-1} \equiv (-1)^{2^m-1} \equiv -1 \pmod{3}$$
or 
$$2^{2^m-1} +1 = \frac{2^{2^m} +2}{2} \equiv 0 \pmod{3}$$
or 
$$\frac{F_m +1}{2}=\frac{2^{2^m} +2}{2} \equiv 0 \pmod{3}$$
Thus, $F_m$ should not necessarily be prime.
