$F_n$ is the $n$-th Fermat Number. Prove there are infinitely many values of n for which $F_n + 2$ is composite.

Question

$F_n$ is the $n$-th Fermat Number. $F_n = 2^{2^n} + 1$. Prove there are infinitely many values of n for which $F_n + 2$ is composite. I tried using reduce modulo 7 but got stuck. Any help is appreciated!

Answer 1

Note that $$F_{n+1}-1=(F_n-1)^2,$$ that is, $$F_{n+1}=(F_n-1)^2+1.$$ Now $$F_0=3,$$ and $$F_n\equiv3\pmod7\implies F_{n+1}\equiv5\pmod7,$$ and $$F_n\equiv5\pmod7\implies F_{n+1}\equiv17\equiv3\pmod7.$$ It follows that $F_n\equiv5\pmod7$ for $n=1,3,5,7,\dots$, and $F_n+2$ is composite for $n=3,5,7,\dots$.

Alternatively, you could use the recurrence $$F_{n+2}=(F_n-1)^4+1.$$