Primes and six

Using just $p$ and $q$

$$\begin{array}{rl}pq+1& =p(p+2)+1\\ & =p^2+2p+1\\ & =(p+1{)}^2\end{array}$$ $p$ is odd so $p+1$ is even

$p,q\gt 3$ so 

$$\begin{gathered} \text{not multiples of 3} \\ \begin{array}{rlrl}& \downarrow & & \downarrow \\ & p& p+1& q\\ & & \uparrow & \end{array} \\ \text{a multiple of 3} \\ \text{(since 1 in every 3 numbers is a multiple of 3)} \end{gathered}$$ So $p+1$ is a multiple of $2$ and $3$

$\therefore p+1$ is a multiple of $6$

$\therefore (p+1)^2$ is a multiple of $36$, and $(p+1)^2=pq+1$

Using another letter $k$

Let $k=p+1$. We note that $k$ is even because any prime $p>3$ is odd.

Moreover, neither $p$ nor $q$ can be divisble by three because the only prime number divisible by three is $3$ itself. On the other hand, if you have three consecutive integers exactly one of them must be divisible by three. Thus, $k$ is also a multiple of $3$.

So, we can write $k$ in the form $k=6n$ for some integer $n$. We deduce further that

$$pq=(6n-1)(6n+1)=36n^2-1$$

which implies that $pq+1 = 36n^2$ is indeed divisible by $36$.