Question

\(P\) and \(Q\) are two positive integers such that \(P^2 = Q^2 + 13\).
The product of the numbers \(P\) and \(Q\) is ______

• A. 13
• B. 26
• C. 39
• D. 42

Hint:

Factor \(P^2-Q^2=13\) as \((P-Q)(P+Q)=13\) and use the fact that 13 is prime.

Answer 1

The Correct Option is D

Instead of factoring, bound $Q$ directly by comparing consecutive squares, since $P^2-Q^2=13$ is a small, fixed gap.

1. Set up the gap: we need two positive integers $P>Q$ whose squares differ by exactly $13$. Consecutive squares grow apart as $(Q+1)^2-Q^2=2Q+1$, so only a small range of $Q$ can give a gap as small as $13$.
2. Try $P=Q+1$: then $P^2-Q^2=2Q+1$. Setting $2Q+1=13$ gives $Q=6$, so $P=7$. Check: $7^2-6^2=49-36=13$. This works.
3. Rule out $P=Q+2$ or larger gaps: if $P=Q+2$, then $P^2-Q^2=4Q+4$, which equals $13$ only if $Q=2.25$, not an integer. Larger gaps between $P$ and $Q$ push $P^2-Q^2$ up even faster relative to $Q$, so no other integer solution exists once $P=Q+1$ already fits exactly.

So the only positive integer solution is $P=7$ and $Q=6$, and their product is $P \times Q = 7 \times 6 = 42$.

The product of the numbers $P$ and $Q$ is $42$, matching option (D).