Find all primes of the form $2^{2^n} + 5$ for a nonnegative integer n

Question

I'm a little lost on how to do this problem. It looks a lot like the definition for the Fermat numbers: $F_n = 2^{2^n} + 1$, however I'm not sure how to use that in order to find all of the primes of the form: $2^{2^n} + 5$.

Answer 1

Hint: Look at $2^{2^n} + 5$ modulo $3$ for $n \geq 1$.

I got this hint by factoring $2^{2^n}+5$ for the first few values of $n$ using the sage command [factor(2^(2^n)+5) for n in range(6)]. When in doubt, experiment!

Answer 2

$$ 2^{2^1}=2^2 \equiv_3 1 $$ $$ 2^{2^2}=(2^2)^2 \equiv_3 1 $$ $$ \dots $$ $$ 2^{2^n}=(2^{2^{n-1}})^2 \equiv_3 1 $$