If $p=2q +1$ with $p,q$ prime then $p$ is called safe prime and $q$ is a Sophie Germain prime. I want a faster algorithm for a safe prime test than doing two primality checks for $p$ and $q$.

In http://en.wikipedia.org/wiki/Safe_prime you find the hint: However, Pocklington's criterion can be used to prove the primality of 2p+1 once one has proven the primality of p.

Are this statement and it's proof correct?

If $p=2q+1$ with $q\,$ prime and $p\,$ passes the Fermat primality test to base $2$, $p \not \equiv 0 \pmod 3$, then $p\,$ is prime.

Proof: We have the following three facts

1) $q$ is prime, $q | p-1,\;$ and $q > \sqrt{p}-1$

2) $2^{p-1}\equiv 1 \pmod p\;$ because $p$ passes the Fermat primality test to base $2$. And

3) $\gcd(2^{(p-1)/q}-1,p)=\gcd(2^2-1,p)= \gcd(3,p) = 1\;$ because $p$ is no multiple of $3$.

Therefore $p$ is prime according to the Pocklington_criterion.

Update: I added the condition $p \not \equiv 0 \pmod 3$ because otherwise the GCD in 3) is not necessarily $=1$. In the practical implementation this is done while sieving for $p$.


1 Answer 1


That $p$ passes the base-$2$ Fermat test means that the order of $2$ modulo $p$ is a divisor of $p - 1 = 2q$. Since $q$ is a prime, we have $p = 2q+1 \geqslant 5 > 2^2$, so the order of $2$ modulo $p$ must be larger than $2$. This leaves the two options $q$ and $2q$ for the order. In either case, $q$ divides the order, so $p$ must have a prime factor $r \equiv 1 \pmod{q}$. Since $q + 1 < p = 2q + 1 < 2(q+1)$, it is impossible that $q+1$ is a prime factor of $p$. Therefore the smallest candidate for $r$ is $r = 2q + 1 = p$. In other words, it follows that $p$ is a prime. (Note we did not need the assumption that $3 \nmid p$ in this argument, that is only needed for a verbatim application of Pocklington's criterion. If $p$ is a multiple of $3$, it won't pass the Fermat test.)


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