Sophie Germain Prime cannot be 6n+1
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Theorem
Let $p$ be a Sophie Germain prime.
Then $p$ cannot be of the form $6 n + 1$, where $n$ is a positive integer.
Proof
Let $p$ be a Sophie Germain prime.
Then, by definition, $2 p + 1$ is prime.
Aiming for a contradiction, suppose $p = 6 n + 1$ for some $n \in \Z_{>0}$.
Then:
\(\ds 2 p + 1\) | \(=\) | \(\ds 2 \paren {6 n + 1} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12 n + 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \paren {4 n + 1}\) |
and so $2 p + 1$ is not prime.
The result follows by Proof by Contradiction.
$\blacksquare$