Given two prime numbers that satisfy the equation $p^{2}-q^{2}=2p-10q+24$, how many pairs of $(p, q)$ are there? I had a seventh grade Math Olympiad, and one specific question stumped me. I memorized this question after the test, here it is:

Given two prime numbers that satisfy the equation $p^{2}-q^{2}=2p-10q+24$, how many pairs of $(p, q)$ are there?

I first thought of the fact that $p \neq q$ since $2p-10q$ could not be $0$ (as opposed to $p^{2}+q^{2}$), but then I realized that it could be possible because of the $+\ 24$.
EDIT: I also converted $p^{2}+q^{2}$ to $(p+q)(p-q)$, but I didn't know where to continue.
I couldn't really think of any tips or tricks to these types of questions as I only had about 4-5 minutes to do this question (the test was 40 minutes long, and there were 10 questions). Would there be any trick to do these types of questions in under 4 minutes or less?
Note that this is a seventh grade Math Olympiad test, so I would accept answers that are AT MOST at a tenth grade level.
 A: Method $-1$
$$\begin{align}&p^{2}-q^{2}=2p-10q+24\\
\iff &(p-1)^2=(q-5)^2 \\
\iff &|p-1|=|q-5| \\
\iff &p-1=±(q-5)\\
\iff &q-p=4 ~\text{or}~ q+p=6 \end{align}$$

Method $-2$
$$\begin{align}&p^{2}-q^{2}=2p-10q+24\\
\implies &p^2-2p-(q^2-10q+24)=0\\
\implies &\Delta=1+q^2-10q+24\\ \implies&\Delta=q^2-10q+25\\
\implies &\Delta=(q-5)^2\\
\implies &p=1±(q-5)\\
\implies &p=q-4 ~\text{or}~ p=6-q\end{align}$$
Method $-3$ (similar)
$$\begin{align}&p^{2}-q^{2}=2p-10q+24\\ 
\implies &q^2-10q-(p^2-2p-24)=0\\
\implies &\Delta =25+p^2-2p-24\\
\implies &\Delta =p^2-2p+1 \\
\implies &\Delta=(p-1)^2\\
\implies &q=5±(p-1)\\
\implies &q-p=4 ~\text{or}~ q+p=6
\end{align}$$
Conclusion:
If $q+p=6$, then $p=q=3$
If $q-p=4$, then this is a specific case of Polignac's conjecture for $n=4$.
A: That's funny.
I just answered this other question:
How to find the smallest $a,b ∈ N$ that solve a single equation
My answer was general enough
to answer this one.
Using my notation there,
$u=-2, v=-10, w=24$
so we look at the factors of
$u^2-v^2+4w
=4-100+96
=0$.
Surprise - I did not consider
this case.
The equation becomes
$(2p-2)^2=(2q-10)^2$
or
$(p-1)^2=(q-5)^2$
or
$0
=(p-1+q-5)(p-1-q+5)
$
so
$0
=p+q-6
$
or
$0
=p-q+4
$.
For the first,
$p+q=6$
so,
if $p, q > 1$,
$p=q=3$.
For the second
$q = p+4$
and there probably
an infinite number of solutions to this
of the form
$p=6m+1, q=6m+5
$.
The first two are
(7, 11), (13, 17).
