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Primes and Six

Age 14 to 16 Short
Challenge Level

Using just $p$ and $q$
$$\begin{split}pq+1&=p(p+2)+1\\
&=p^2+2p+1\\
&=(p+1)^2\end{split}$$ $p$ is odd so $p+1$ is even
$p,q\gt 3$ so 
$$\text{not multiples of 3}\\
\begin{align} &\downarrow & &\downarrow\\
&p &p+1\hspace{5mm} &q\\
& &\uparrow\hspace{10mm}&  \end{align}\\
\text{a multiple of 3}\\
\text{(since 1 in every 3 numbers is a multiple of 3)}$$ 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 $36$.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.