Elementary Number Theory : Prime Problem : Determine all twin primes $p$ and $p+2$ for which $p(p+2)-2$ is also prime.
First, $3$ and $5$ are primes and $3*5-2=13$ is also prime.
Assume that $p=3m+n$ $(n=0, 1, 2)$ and 1$\leq$m. i.e. $p>3$
Then $n$ cannot be $0$ since $p$ is prime.
And $p$ cannot be $3n+1$ since $(3n+1)+2=3(n+1)$ is not prime.
So $p=3n+2$
Then $p(p+2)-2=(3n+2)(3n+4)-2=3(3m^2+6m+2)$ is not prime.
Hence $3, 5$ are the only pair.
Is this right proof?

Apart from a mistyping (you said $3×5=15$ where you meant $3×5-2=13$) the proof looks correct.
An alternate version: If $p\equiv2\bmod3$ then we can render
$p(p+2)-2\equiv2(2+2)-2\equiv6\equiv0\bmod3$
without computing the full polynomial.
