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:

Write P squared minus Q squared as a difference of squares and use that 13 is a prime number to fix P minus Q and P plus Q.

Answer 1

The Correct Option is D

A quicker route is to guess that $P$ and $Q$ are close together, since their squares differ by only $13$, a small number. The simplest guess is that they are consecutive integers, $P=Q+1$.

Substitute this into the given equation:

\[ (Q+1)^2=Q^2+13 \]

Expand the left side:

\[ Q^2+2Q+1=Q^2+13 \]

The $Q^2$ terms cancel, leaving:

\[ 2Q+1=13 \implies 2Q=12 \implies Q=6 \]

So $P=Q+1=7$.

This guess is not just lucky, it is forced: since $13$ is prime, the factor pair $(P-Q, P+Q)$ of $13$ can only be $(1,13)$, so $P-Q$ must equal exactly $1$, which is precisely the consecutive-integer assumption used here. So $P=7$ and $Q=6$ is the only solution in positive integers.

The required product is:

\[ P\times Q=7\times6=42 \] \[ \boxed{42} \]