Is this proof acceptable ?


Let $\displaystyle p$ be prime greater than $\displaystyle 3 $.
​If $\displaystyle 2p+1 $ is prime then $\displaystyle 2p+1 \mid \frac{3^p-1}{2}$.


Suppose $\displaystyle q=2p+1$ is prime. $\displaystyle q \equiv 11 \pmod{12}$ so $\displaystyle 3$ is quadratic residue module $\displaystyle q$
​and it follows that there is an integer $\displaystyle n$ such that $\displaystyle n^2 \equiv 3 \pmod{q}$ . This shows
$\displaystyle 3^p=3^{(q-1)/2} \equiv n^{q-1} \equiv 1 \pmod{q}$ showing $\displaystyle 2p+1 $ divides $\displaystyle 3^p-1$ ,
​and since $\displaystyle 2p+1$ is odd number it divides $\displaystyle \frac{3^p-1}{2}$ .