How to prove $2^{2^{6n+2}} + 21$ is a composite number?

Question

How to prove that $$ 2^{2^{6n+2}} + 21 $$ is a composite number for each $$n=1,2,3,\ldots$$

I'm stuck on a what seems like a simple problem but I don't even know where to start.

Answer 1

Hint: Try reducing modulo some small primes.

Answer 2

Let's denote the number by $F(n)$, for $n=0,1,2,...$
One can prove by induction that $37$ divides $F(n)$ for all $n$.
This is easy to see if one uses Fermat's little theorem.
I have certain problems with MathJax so I cannot write the full proof here.

EDIT:

1) For $n=0$, we have $F(0) = 37$ which is divisible by 37.
2) For $F(n)$ we assume that $37$ divides $F(n)$.
3) Now for $F(n+1)$ we have:
$F(n+1) = 2^{2^{6n+8}} + 21 = {(2^{2^{6n+2}})}^{2^6} + 21 = (mod\ 37) = 16^{2^6} + 21 = 16^{64} + 21 = 2^{256} + 21 = 2^{(36.7)}.2^4 + 21 = {2^{36}}^7.2^4 + 21 = 1^7.16 + 21 = 37 = 0\ (mod \ 37)$

Answer 3

$ 2^{2^{6n+2}} + 21=2^{4 (2^{6n} )}-2^4+2^4+21= $

$ =37+2^4\left (2^{4(2^{6n}-1 )}-1 \right )=37+2^4\left (2^{4(2^6-1 )(2^{6(n-1)}+ \dots +1 )}-1 \right )=$

$ =37+2^4\left (2^{36 \cdot 7 \cdot (2^{6(n-1)}+ \dots +1 )}-1 \right )=$

$ =37+2^4\left (2^{36}-1\right )\left (\dots \right ) \equiv 0 \bmod 37$