On a result of Euler on pseudoprimes 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...
 A: 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.
