$2^{p}\equiv1$ mod $2p+1$ for certain $p$

Question

Let $p$ be a prime number such that $p\equiv3$ mod $4$ and $2p+1$ is also a prime number. It is well known that $2^{p}\equiv1$ mod $2p+1$, but I haven't been able to prove it.

Answer 1

By Fermat's theorem $$2^{2p} \equiv 1 \pmod{2p+1}.$$ Then using the prime property we can say that $2p+1$ either divides $2^p-1$ or $2^p+1$, i.e. at least one of the following is true: $$2^p \equiv 1 \pmod{2p+1} \qquad \qquad \text{or} \qquad \qquad 2^p \equiv -1 \pmod{2p+1}.$$ Observe that both these congruences cannot be true together otherwise $2p+1$ will divide $2$. So we need to rule out that the second congruence should not hold.

Here is a hint: Now use the fact that $p \equiv 3 \pmod{4}$ and think in terms of quadratic residues. Let me know if you need more explanation.