Suppose there is such $k$ and let $t$ be the order of $2$ mod $2p+1$. Then $t|2p+1-1=2p$ and $t|k$ so in particular $t<2p$. This implies that $t=p$ since, as one can easily verify, $t$ can't be $1$ or $2$. But then we also have: $2^{p+1} \equiv 2 \mod{2p+1}$ and since $p+1$ is even it follows that $2$ is a quadratic residue mod $2p+1$ which is impossible because $2p+1 \equiv 3 \mod{8}$ and $2$ is a quadratic residue only for primes which are $1$ or $7 \mod{8}$.

Peter wrote: Let $p$ be a prime number of the form $4k+1$. Suppose that $2p+1$ is prime. Show that there is no $k \in \mathbb{N}$ with $k<2p$ and $2^k \equiv 1 \; \pmod{2p+1}$. I think that $p$ must be a prime in the form of $4r+1$ instead of $4k+1.$

crazyfehmy wrote: Peter wrote: Let $p$ be a prime number of the form $4k+1$. Suppose that $2p+1$ is prime. Show that there is no $k \in \mathbb{N}$ with $k<2p$ and $2^k \equiv 1 \; \pmod{2p+1}$. I think that $p$ must be a prime in the form of $4r+1$ instead of $4k+1.$ Yes,I think so