Is this proof acceptable ?


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


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