Prove by induction that $2^{2n} – 1$ is divisible by $3$ whenever n is a positive integer. I am confused as to how to solve this question.
For the Base case $n=1$, $(2^{2(1)} - 1)\,/\, 3 = 1$, base case holds
My induction hypothesis is:
Assume $2^{2k} -1$ is divisible by $3$ when $k$ is a positive integer
So, $2^{2k} -1 = 3m$ 
$2^{2k} = 3m+1$
after this, I'm not quite sure where to go.  Can anyone provide any hints? 

If we rewrite $2^{2k}-1$ in binary, we get the number 
$$2^{2k}-1=\underset{\text{$2k$-times}}{\underbrace{111111\dots11}}$$
consisting of $2k$ ones. (I.e., there are $k$ pairs of them.)
If this is not clear immediately, just notice that by adding one to the above number we get
$$2^{2k}=1\underset{\text{$2k$-times}}{\underbrace{000000\dots00}}.$$
Now we can simply notice that this number can be obtained as the sum (written in binary)
$$
\begin{align*}
110000\dots00&+\\
  1100\dots00&+\\
        \dots&+\\
           11&=\\
111111\dots11&
\end{align*}
$$
where each summand is multiple of $3=(11)_2$.
