How does one show that for $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd. 
For $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd. 

Firstly,
$k \geq 1$
I can see induction is the best idea:
Show for $k=1$:
$2^{2^1} + 5 = 9 , 2^{2^1} + 3 = 7$
Assume for $k = \mu$
so: $3\mid2^{2^\mu} + 5 , \space 7\mid2^{2^\mu} + 3$
Show for $\mu +2$
Now can anyone give me a hint to go from here? My problem is being able to show that $2^{2^{\mu+2}}$ is divisible by 3, I can't think of how to begin how to show this. 

We have $2^2\equiv 1\pmod 3$ now one notes that $$2^{2^k}=(2^2)^{2^{k-1}}\equiv 1\pmod 3$$ so $\forall k$ we have $$2^{2^k}+5\equiv 6\equiv 0\pmod 3$$
We also have $2^{2^2}\equiv 2\pmod 7$ this means $2^{2^k}\equiv 2^{2^{k-2}}\pmod 7$ So when $k$ is odd $2^{2^k}\equiv 2^2\pmod 7$ and therefore $2^{2^k}+3\equiv 0\pmod 7$
