Proving that $2^{2^n} + 5$ is always composite by working modulo $3$ By working modulo 3, prove that $2^{2^n} + 5$ is always composite for every positive integer n.
No need for a formal proof by induction, just the basic idea will be great. 

In order to work out this problem, we start by noticing $2^2\equiv 1 ~~~(\text{mod } 3)$. What this means is $2^2$ (which is $4$) is $1$ greater than some multiple of $3$ (that multiple, in this case is obviously, $3$)
So if $2^2\equiv 1 ~~~(\text{mod } 3)~~\implies (2^2)^n\equiv (1)^n ~~~(\text{mod } 3)\implies 2^{2n}\equiv 1~~(\text{mod } 3)$,
Again, this means that any even power of $2$ is $1$ greater than some multiple of $3$. So if $2^{2n}$ is $1$ greater than some multiple of $3$ (another way to say this is that $2^{2n}$ leaves a remainder of $1$ when divided by $3$), then what can you say about $2^{2n}+5$?.

To answer this, forget about $2^{2n}$ and just think about the remainder (i.e $1$). This is how modular arithmetic makes our live a lot easier. If $2^{2n}$ is $1$ greater than some multiple of $3$, then $2^{2n}+5$ should be $1+5=6$ greater than that multiple of $3$, right? But you know that $6$ is, by itself, a multiple of $3$. So, this should mean that $3|2^{2n}+5$.
A more formal (and a neat) argument could be: 
$3|2^{2n}-1\implies 3|2^{2n}+5$, since $2^{2n}+5=(2^{2n}-1+6)$


Now since it has been shown that $2^{2n}+5$ is infact a multiple of $3$, it should be apparent that $2^{2n}+5$ is after all, composite.

I Hope it helps!
