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In several sources (for instance on page 58 of the first ed. of Crandall & Pomerance book on prime numbers or at the end of this paper by J. H. Jaroma), I have seen a result that goes like this:

Let $p$ be an odd prime congruent to $-1$ modulo $4$. Then $2p+1$ is a prime iff $(2p+1) \mid (2^{2p}-1)$.

Do you know if the hypothesis $p \equiv -1 \pmod 4$ may be removed? If the result is still valid for primes congruent to $1$ modulo $4$, I wonder why it is that it is not mentioned in the sources I referred to above...

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    $\begingroup$ This post on Mathematics seems related: Proving $2p +1 \mid 2^p + 1$. $\endgroup$ Feb 13, 2021 at 13:27
  • $\begingroup$ Thanks, Martin... The result there allows us to prove this: if p is congruent to 1 mod. 4 and $2p+1$ is a prime then 2p+1 divides $2^{2p}-1$. $\endgroup$
    – Jamai-Con
    Feb 13, 2021 at 13:37
  • $\begingroup$ Hmm, thus this is true for all primes including the even prime (p=2). $\endgroup$
    – Wlod AA
    Jul 14, 2021 at 7:26

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Yes, the result holds for every odd prime number $p$... I certainly find it somewhat "strange" that it is only stated for primes congruent to $-1$ modulo $4$ in several places:

Proposition. Let us suppose that $p$ is an odd prime number and that $2p+1$ divides $2^{2p}-1$. Then, $2p+1$ is a prime number.

Proof. (I learnt it from J. I. Restrepo) For the sake of contradiction, let us suppose that $2p+1$ is not a prime number and that $q$ is a prime number dividing $2p+1$. From the hypothesis and Fermat's little theorem we have that

\begin{eqnarray} 2^{2p} \equiv 1 \pmod{q}\\ 2^{q-1} \equiv 1 \pmod{q}. \end{eqnarray}

It follows from these congruences that $\mathrm{ord}_{q}(2)=:\mathfrak{o}$ is a common positive divisor of $2p$ and $q-1$; since there are only four positive divisors of $2p$ and $q-1 \leq (p-\frac{1}{2})<p$, we get that $\mathfrak{o}=1$ or $\mathfrak{o}=2$. Given that $q$ is a prime number, $\mathfrak{o}$ can't be equal to $1$. Hence, $\mathfrak{o}=2$ and $q=3$.

From what we have established in the above paragraph, we obtain that $2p+1=3^{\ell}$ for some $\ell \in \mathbb{Z}^{+} \setminus \{1\}$. This implies that $9 \mid (2^{2p}-1)$ which is an absurdity because $9 \mid (2^{n}-1)$ iff $6 \mid n$. Q.E.D.

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