Prove that there are infinitely many composite numbers of the form $2^{2^n}+3$. 
There are infinitely many composite numbers of the form $2^{2^n}+3$.
  [Hint: Use the fact that $2^{2n}=3k+1$ for some $k$ to establish that $7\mid2^{2^{2n+1}}+3$.] 

If $p$ is a prime divisor of $2^{2^n}+3$ then $2^{2^n}+3\equiv0\pmod{p}$. But I don't think this is useful. I don't see how the fact that $2^{2n}=3k+1$ helps to show that $7\mid2^{2^{2n+1}}+3$.
Any hints on how to start solving this problem?

Let $p$ be a prime. Then by Fermat's little theorem, $2^{p-1} \equiv 1 \pmod{p}$, so $2^{k(p-1) +2} \equiv 4 \pmod p$. So we wish to find conditions on $n$ so that $2^n \equiv 2 \pmod{p-1}$.  Now for $p=7$, $p-1 = 6$ and the sequence $2^n$ modulo $6$ looks like this, starting with $n=1$ : 
$$
2, 4, 2, 4, 2, 4, \cdots
$$
so for $n=2m+1$, $2^{2m+1} \equiv 2 \pmod 6$, which means 
$$
2^{2^n}+3 \equiv 2^{2^{2m+1}} + 3 \equiv 2^2+3 \equiv 0 \pmod 7. 
$$ 
You should have just used the hint! But you asked the question here so I figured it didn't tell you much. 
P.S. : If I used the hint a 100%, I would have written $2^{2m} = 3k+1$ which implies that $2^{2m+1} = 6k+2$, which is essentially my remark that $2^{2m+1} \equiv 2 \pmod 6$. 
Hope that helps,
