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We want to show whether or not $2^{2n+1}+1$ Is divisible by 3 using induction.

I tried doing this but I'm stuck with $2^{2n+3} + 1= 2^2 \cdot 2^{2n+1} + 1$. I don't know how to show if this is divisible by 3.

Johannes
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    $2^{2n+3} + 1= 2^2 \cdot 2^{2n+1} + 1 = 3\cdot 2^{2n+1} + 2^{2n+1} + 1$ and you are through. –  Sep 06 '16 at 11:03

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For $n=0$ this should be clear. So suppose we know that $2^{2\cdot n + 1}$ is divisible by $3$. Then

\begin{align} 2^{2\cdot (n +1) + 1} &= 2^{2\cdot 1 + 2\cdot n + 1}\\ &= 2^2 \cdot 2^{2\cdot n + 1} \end{align} is also divisible by $3$, since the second factor is.

Kaligule
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