What are all the twin primes $p$ and $q = p + 2$ for which $pq - 2$ is also prime? It seems that $p = 3$ and $q = 5$ ($pq - 2 = 13$) are the only solutions. However I'm having a difficult time proving this. 
I have that all primes can be represented as $3k + 1$ or $3k + 2$ so if $p = 3k + 2$ then $q = 3k + 4$ and 
\begin{align*}
  pq - 2 &= (3k + 2)(3k + 4) - 2\\
     &= 9k^2 + 18k + 8 - 2\\
     &= 9k^2 + 18k + 6\\
     &= 3(3k^2 + 6k + 2) \text{             which can't be prime}
\end{align*}
So that seems promising. But when I let $p = 3k + 1$ and $q = 3k + 3$ I get 
\begin{align*}
pq -2 &= (3k + 1)(3k + 3) - 2\\
&= 9k^2 + 12k + 3 - 1\\
&= 9k^2 + 12k + 2,
\end{align*}
which could very well be prime. So what do I do? It appears, though, that the only solution is $p = 3$. 

We notice that for $p=3$ and $q=5$ we have a solution. 
Let $p \geq 5$ 
Then p has form $p=6k \pm 1$ where $k \in N$
If $p=6k+1$ then $q=6k+3=3(2k+1)$ so $q$ is not a prime then. That leaves us with $p=6k-1$. Then $q=6k+1$ so $pq-2=(6k-1)(6k+1)-2=36k^2-3=3(12k^2-1)$ which is not a prime.
We conclude that there is no solution for $p \geq 5$ so the only solution is $p,q (3,5)$
