$p$ is a prime greater than 3. If $d\mid\frac{2^p+1}{3}$ then $p\mid d-1$

Question

$p$ is a prime greater than 3. If $d \mid \dfrac{2^p+1}{3}$. Prove that $p \mid d-1$. I have no clue how to solve this

Answer 1

It is sufficient to prove that for each prime divisor $q$ of $d$, we have $p \mid q-1$. We have $4^p=1$ mod $q$, so the order of 4 in $\mathbb{Z}_q^*$ is equal to $p$; thus $p$ divides $q-1$.