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 $p = 3$. 

Hint $\,\ {\rm mod\ 3}\!:\ p\not\equiv \color{#c00}0\,\Rightarrow\, \color{#0a0}q\,(pq\!-\!2) = \overbrace{(p\,+\,2)}^{\large 0\ {\rm  if}\ p\,\equiv\,\color{#c00}{\bf 1}\ }\,\overbrace{(p\,(p+2)-2)}^{\large\ 0\ {\rm if}\ p\,\equiv\,\color{#c00} 2}\equiv 0 $ 
hence we conclude  $\ 3\nmid \color{#0a0}q\,\Rightarrow\,3\mid pq-2$
