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. 

Hint $\,\ $ Apply the following $ $ fixed point induction $ $  to $\ f(j)\, =\, 2^{\Large 2^{\Large 2j+1}}$
Lemma $\ $ If $\,f(j\!+\!1)\color{#c0f} = f(j)^4\,$ then $\ {\rm mod\ n}\!:\,\ \forall n\ge 1\!:\ f(n)\color{#0a0}\equiv f(1) \Leftarrow\!\Rightarrow f(1)^4\color{#c00}\equiv f(1)\,$
Proof $\quad\ (\Rightarrow)\quad f(1)^4\color{#c0f}\equiv f(2)\color{#0a0}\equiv f(1).\quad\ (\Leftarrow)\quad f(n+1) \equiv f(n)^4\overset{induct}\equiv f(1)^4\color{#c00}\equiv f(1)$
Remark $\ $ So the gist is: successive iterations stay constant iff the initial point is fixed.
