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

Step 1: Write the given relation as a difference of squares.
We are told \(P^2 = Q^2 + 13\), so \(P^2 - Q^2 = 13\). Using the identity \(a^2 - b^2 = (a-b)(a+b)\), this becomes
\[ (P-Q)(P+Q) = 13 \]

Step 2: Use the fact that \(13\) is prime.
\(P\) and \(Q\) are positive integers, so \(P-Q\) and \(P+Q\) are integers, and \(P+Q>0\). Since \(P^2>Q^2\), we must have \(P>Q\), so \(P-Q\) is a positive integer too. Both factors are positive integers whose product is \(13\), a prime number, so the only way to split \(13\) into two positive integer factors is \(1 \times 13\). So
\[ P - Q = 1, \qquad P + Q = 13 \]

Step 3: Solve the two equations for \(P\) and \(Q\).
Adding the two equations: \[ (P-Q) + (P+Q) = 1 + 13 \implies 2P = 14 \implies P = 7 \]
Putting this back into \(P+Q=13\) gives \(Q = 6\). Check: \(P^2 = 49\) and \(Q^2+13 = 36+13 = 49\), which matches, confirming \(P=7\), \(Q=6\) is correct.

Step 4: Find the product.
\[ P \times Q = 7 \times 6 = 42 \]

Final Answer:
The product of \(P\) and \(Q\) is \(42\), so option (D) is correct. \[ \boxed{PQ = 42} \]