Primes and Six
Weekly Problem 1 - 2015
If $p$ and $q$ are prime numbers greater than $3$ and $q=p+2$, prove that $pq+1$ is divisible by $36$.
If $p$ and $q$ are prime numbers greater than $3$ and $q=p+2$, prove that $pq+1$ is divisible by $36$.
Problem
Let $p$ and $q$ be prime numbers with $q=p+2$ and $p$ greater than $3$.
Prove that $pq+1$ is divisible by $36$.
If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.
Student Solutions
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 $36$.