Difference between Sophie Germain primes are divisible by 3?

Question

It seems that the difference between two Sophie Germain primes greater than 3 is divisible by 3. If this is true, how to prove it?

A Sophie Germain prime is a prime $p$ such that $2p+1$ also is a prime.

Answer 1

A prime $>3$ is congruent to one of $\pm1\pmod6$. If $p\equiv1\pmod6$ then $2p+1\equiv3\pmod6$ is divisible by three and therefore $2p+1$ is not a prime.

So if $p$ is the smaller member of a Sophie Germain pair, we must have $p\equiv-1\pmod6$. Then $2p+1\equiv-1\pmod6$ as well, and the difference $p+1$ is divisible by six.

Answer 2

For $2p+1$ to be prime you need it to be not divisible by $3$. Thus $p$ must not be $1$ modulo $3$ and thus is $2$ modulo $3$ (except for $p=3$ which you excluded).

Thus every such $p$ is $2$ modulo $3$ and the difference of two such primes is divisible by $3$.