If $F_m = 2^{2^m} + 1$ is prime (with $m \geq 1$), does it follow that $3 \mid \frac{F_m + 1}{2}$?

Question

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 $1 \leq m \leq 4$.

Answer 1

$F_m=2^{2^m}+1$ 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.

Answer 2

For any even $k>0$, $2^k\equiv 1\bmod 3$. So $3 \mid 2^k{+}2$ (and of course $3 \mid \frac{2^k{+}2}{2}$ also), regardless of whether $2^k{+}1$ is prime or not.