<s>[INST] How to prove that all odd powers of two add one are multiples of three 
For example
  \begin{align} 
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}

I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
 [/INST] One of your examples is that $2^{11} + 1$ is divisible by $3$. We investigate as follows:
Let us consider instead raising $2$ to an even power and subtracting $1$. And then let us factor.
Example: $2^{10} - 1 = (2^5 - 1)(2^5 + 1)$.
Among any three consecutive integers, exactly one of them must be divisible by $3$.
Clearly $2^5$ is not divisible by $3$, so either its predecessor or successor is divisible by $3$.
That is, either $2^5 - 1$ or $2^5 + 1$ is divisible by $3$, whence their product is, as well.
Okay: Their product is $2^{10} - 1$, which we have now established is divisible by $3$.
This number is still divisible by $3$ after being doubled, and still divisible by $3$ when we add $3$ to it. 
So: We have that $2(2^{10} - 1) + 3 = 2^{11} - 2 + 3 = 2^{11} + 1$ is divisible by $3$ as desired.
A similar bit of reasoning around $2^{2k} - 1$ yields the assertion at hand. "QED"
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