Show that $2^{2^n}+5$ is divisible by $3$ for all $n \geq 1.$ This is part of a larger proof that I am trying to do, but this is a fact I'm trying to wrap my head around and prove. Based on a Sage program, I was able to deduce by pattern that $2^{2^n}+5$ is divisible by $3$ for all $n \geq 1.$ If you factor numbers of this form until $n=6$ on Sage, you get that a prime factor of all of them is $3.$ I tried proving this through induction, but I couldn't quite figure out how to make the argument for the inductive step. I know modular arithmetic is the way to go, but it isn't something we've studied yet in class! Thank you for your help.

It's easy to prove by induction that $2^m \equiv 1 \pmod 3$ if $m$ is even, because $4 \equiv 1 \pmod 3$ (establishing your base case) and $2^{2(k+1)}=4 \cdot 2^{2k} \equiv (\text{using your inductive hypothesis})~1 \cdot 1 \equiv 1 \pmod 3$.  (Also, $2^m \equiv 2 \pmod 3$ if $m$ is odd, but you don't need that.)  $2^n$ is always even for $n \geq 1$, so $2^{2^n}+5 \equiv 1+5 \equiv 0 \pmod 3$.
