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?

$2 = 3-1$
$2^k = (3 - 1)^k = 3^k - k*3^{k - 1} ..... = \sum_{n = 0}^k {k \choose n}3^{k - n}(-1)^n = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n + {k \choose 3}3^{k - n}(-1)^k = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n \pm 1$
Since $k$ is odd:
$2^k = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n - 1$
$2^k + 1 = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n = 3\sum_{n = 0}^{k - 1} {k \choose n}3^{k - n - 1}(-1)^n$
which is a multiple of 3.
