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

Step 1: Rearrange the equation as a difference of squares.
We are given
\[ P^2=Q^2+13 \]
which rearranges to
\[ P^2-Q^2=13 \]
Using the identity \(a^2-b^2=(a-b)(a+b)\), this factors as
\[ (P-Q)(P+Q)=13 \]
Step 2: Use the fact that 13 is prime.
Since \(P\) and \(Q\) are positive integers and \(P^2>Q^2\), we know \(P>Q\), so \(P-Q\) is a positive integer, and \(P+Q\) is also a positive integer, larger than \(P-Q\). Their product is 13, a prime number, so the only way to write 13 as a product of two positive integers is \(1\times13\). Because \(P+Q>P-Q\), we must assign
\[ P-Q=1, \qquad P+Q=13 \]
Step 3: Solve the two equations together.
Adding the two equations,
\[ 2P=14 \implies P=7 \]
Subtracting the first from the second,
\[ 2Q=12 \implies Q=6 \]
Step 4: Verify the solution.
Check \(P^2-Q^2=49-36=13\), which matches the given equation, so \(P=7\) and \(Q=6\) are correct.
Step 5: Compute the required product.
\[ P\times Q=7\times6=42 \]
Step 6: Final Answer.
The value 13 in option (A) is only the given difference of squares, not the product asked for. Options (B) and (C), 26 and 39, are just 13 multiplied by 2 and by 3, with no basis in the actual solved values of P and Q. The correct product of the two integers is
\[ \boxed{42} \]