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? 

Hint: If $2^{2k} - 1$ is divisible by $3$, then write
\begin{align*}
2^{2(k + 1)} - 1 &= 2^{2k + 2} - 1 \\
&= 4 \cdot 2^{2k} - 1 \\
&= 4 \cdot \Big(2^{2k} - 1\Big) + 3
\end{align*}
Do you see how to finish it up?

This technique is motivated by attempting to shoehorn in the term $2^{2k} - 1$, since that's the only piece we really know anything meaningful about.
