How to prove that all odd powers of two add one are multiples of three For example
  \begin{align} 
2^5 + 1 &amp;= 33\\
2^{11} + 1 &amp;= 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?
Since $2 \equiv -1 \pmod{3}$, therefore $2^{k} \equiv (-1)^k \pmod{3}$. When $k$ is odd this becomes $2^k \equiv -1 \pmod{3}$. Thus $2^k+1 \equiv 0 \pmod{3}$.
There is exactly one integer that divides three among any three consecutive integers. Since $2^{61}-1$ is prime, $2^{61}$ trivially is not a multiple of $3$, $2^{61}+1$ is a multiple of $3$.