Show that a Sophie Germain prime $p$ is of the form $6k - 1$ for $p > 3$ A Sophie Germain prime is a prime $p$ such that $2p + 1$ is also prime. 
According to a comment in OEIS A023212 (https://oeis.org/A023212), such a prime $p$ is of the form $6k - 1$ for $p > 3$.
How would one show that?

Every prime larger than $3$ must be of the form $6k\pm 1$.
(why? Hint: $6k$ and $6k\pm 2$ are even and $6k+3$ is divisible by $3$).
For contradiction, let a Sophie Germain prime $p>3$ be of the form $6k+1$.
But then $2p+1=2(6k+1)+1=3(4k+1)$,
so $2p+1$ is divisible by $3$, so $2p+1=3$, so $p=1$, contradiction.
