# Fermat pseudoprimes p to base 2 (AKA Sarrus or Poulet numbers) with special properties

Are there any known Fermat pseudoprimes $p\;$ to base $2\;$ (Sarrus or Poulet numbers) with the properties $q = (p-1)/2\;$ is prime and $p \equiv 0 \pmod 3?$

I was not able to find any example up to $p = 2^{31}-1.$ Is there an argument that they cannot exist?

This question is related to Fast check of safe primes or Sophie Germain primes where I added the condition $p \not \equiv 0 \pmod 3?$

• Do you know that there's a database of all pseudoprimes up to $2^{64}$ at cecm.sfu.ca/Pseudoprimes ? – Gerry Myerson Jul 22 '14 at 9:25
• @Gerry Myerson: Yes, I know. But this is not a bottleneck. I can generate the $q$ primes and check the conditions, $2^{31}$ was only a limit for quick calculations because behind it I had to use multi-precision arithmetic. – gammatester Jul 22 '14 at 9:34
• I wonder if there's anything to say about divisibility by $2q+1$ for prime $q$. Perhaps is there an interesting group of order $2q+1$ such as we have for $q^k-1$? – punctured dusk Jul 22 '14 at 11:02

No, there can't be Fermat pseudoprimes $p$ to base $2$ with $(p-1)/2$ prime. Proof, detailing Fedor Petrov's answer:

Assume $p=2q+1$ with $q$ prime and $2^{p-1}\equiv1\pmod p$.

For all $x$ coprime with $p$, it holds that $x^{\varphi(p)}\equiv1\pmod p$, where $\varphi$ is Euler's totient function. $p$ is odd, thus holds for $x=2$. Thus $2^{\varphi(p)}\equiv1\pmod p$.

Let $d=\gcd(\varphi(p),p-1)$. By Bézout's lemma, $d=a\;\varphi(p)+b(p-1)$ for some $(a,b)\in\Bbb Z^2$. Therefore, and given that $2$ is coprime with $p$ \begin{align}2^d&\equiv2^{a\;\varphi(p)+b(p-1)}&\pmod p\\ &\equiv{(2^{\varphi(p)})}^a\;{(2^{p-1})}^b&\pmod p\\ &\equiv1^a\;1^b&\pmod p\\ &\equiv1&\pmod p \end{align}

$p$ is odd, thus both $\varphi(p)$ and $p-1$ are even. $d$ divides both, thus is even. $p-1=2q$ with $q$ prime, thus an even $d$ dividing $p-1$ is either $2$ or $p-1$. With $2^d\equiv1\pmod p$, $d=2$ would imply that $p$ divides $3$, which can't hold for $p=2q+1$ with $q$ prime. Therefore, $d=p-1$.

Therefore, $p-1$ divides $\varphi(p)$. Therefore, $p$ is prime.

Note: we did not need to invoke the question's $p\equiv 0\pmod3$ or the Pocklington criterion.

• Thank you for the question and link to mathoverflow. Here $n$ should be replaced by $p$. – gammatester Mar 19 '18 at 9:11
• @gammatester: done. Thanks for introducing the question, which had been puzzling me for weeks in that context (though not as long as you!). – fgrieu Mar 19 '18 at 9:38
• Oh, I had missed that earlier equivalent proof by Daniel Fischer – fgrieu Mar 19 '18 at 14:48