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.

• There are two squares there waiting to be completed.
– dxiv
Jul 17 '21 at 3:35
• Write it as $p^2-2p = q^2-10q+24$. Jul 17 '21 at 3:40
• There's a problem with the question. From $(p-1)^2 = (q-5)^2$, this implies either $p-1 = q-5$ or $p-1 = -(q-5)$. But in the first case, there are many, many solutions: see cousin prime on Wikipedia. Jul 17 '21 at 3:49
• @TobyMak Good point. Hard to imagine that a 7th grader is expected to be able to prove, on-the-fly that there are an infinite number of cousin primes. Jul 17 '21 at 3:54
• @TobyMak No doubt the problem composer is from the future and the composer alone knows that the person who will prove the cousin prime conjecture is one of the contestants. Surely you have watched the Star Trek movies. Jul 17 '21 at 3:58

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 $$n=4$$.

• Thank you for this answer! +1 and correct answer marked. Jul 17 '21 at 4:45
• @Tyrcnex You're welcome. Actually, I don't think I'm very helpful. Because, I did not expect to encounter an open problem at the end. Jul 17 '21 at 4:56
• Well, at least now I know that the people who put this question in didn't know what they were doing :) Jul 17 '21 at 5:04

That's funny. I just answered this other question: How to find the smallest $a,b ∈ N$ that solve a single equation

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 $$p=6m+1, q=6m+5$$.