Question:
\(P\) and \(Q\) are two positive integers such that \(P^2 = Q^2 + 13\).
The product of the numbers \(P\) and \(Q\) is ______
\(P\) and \(Q\) are two positive integers such that \(P^2 = Q^2 + 13\).
The product of the numbers \(P\) and \(Q\) is ______
The product of the numbers \(P\) and \(Q\) is ______
Show Hint
Rewrite the equation as a difference of squares, (P-Q)(P+Q) = 13.
Since 13 is prime, find the two factors and solve for P and Q.
Since 13 is prime, find the two factors and solve for P and Q.
Updated On: Aug 5, 2026
- 13
- 26
- 39
- 42
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Understand the question:
We are told that \(P\) and \(Q\) are positive integers and \(P^2 = Q^2 + 13\), and we need to find the actual values of \(P\) and \(Q\) so we can compute their product.
Step 2: Rearrange into a difference of squares.
Moving \(Q^2\) to the left side gives \(P^2 - Q^2 = 13\). Using the identity \(a^2 - b^2 = (a-b)(a+b)\), this becomes:
\[ (P - Q)(P + Q) = 13 \]
Step 3: Use the fact that 13 is prime.
Since 13 is a prime number, its only positive factor pairs are 1 and 13. Because \(P\) and \(Q\) are positive, \(P + Q\) must be bigger than \(P - Q\), so we must have \(P - Q = 1\) and \(P + Q = 13\).
Step 4: Solve for P and Q.
Adding the two equations gives \(2P = 14\), so \(P = 7\). Putting \(P = 7\) back into \(P - Q = 1\) gives \(Q = 6\). Checking, \(7^2 = 49\) and \(6^2 + 13 = 36 + 13 = 49\), so this fits the original equation.
\[ P \times Q = 7 \times 6 = 42 \]
Step 5: Check option (A) 13.
13 is the difference \(P^2 - Q^2\), not the product \(P \times Q\), so mixing these up gives this wrong option.
Step 6: Check option (B) 26.
26 is twice 13, which does not come from the actual values \(P = 7\) and \(Q = 6\), so this option is wrong.
Step 7: Check option (C) 39.
39 is three times 13, again not related to \(P \times Q\), so this option is wrong.
Step 8: Check option (D) 42.
This matches \(7 \times 6\) found in Step 4, so this option is correct.
Final Answer:
The two integers are 7 and 6, so their product is 42. \[ \boxed{P \times Q = 42} \]
We are told that \(P\) and \(Q\) are positive integers and \(P^2 = Q^2 + 13\), and we need to find the actual values of \(P\) and \(Q\) so we can compute their product.
Step 2: Rearrange into a difference of squares.
Moving \(Q^2\) to the left side gives \(P^2 - Q^2 = 13\). Using the identity \(a^2 - b^2 = (a-b)(a+b)\), this becomes:
\[ (P - Q)(P + Q) = 13 \]
Step 3: Use the fact that 13 is prime.
Since 13 is a prime number, its only positive factor pairs are 1 and 13. Because \(P\) and \(Q\) are positive, \(P + Q\) must be bigger than \(P - Q\), so we must have \(P - Q = 1\) and \(P + Q = 13\).
Step 4: Solve for P and Q.
Adding the two equations gives \(2P = 14\), so \(P = 7\). Putting \(P = 7\) back into \(P - Q = 1\) gives \(Q = 6\). Checking, \(7^2 = 49\) and \(6^2 + 13 = 36 + 13 = 49\), so this fits the original equation.
\[ P \times Q = 7 \times 6 = 42 \]
Step 5: Check option (A) 13.
13 is the difference \(P^2 - Q^2\), not the product \(P \times Q\), so mixing these up gives this wrong option.
Step 6: Check option (B) 26.
26 is twice 13, which does not come from the actual values \(P = 7\) and \(Q = 6\), so this option is wrong.
Step 7: Check option (C) 39.
39 is three times 13, again not related to \(P \times Q\), so this option is wrong.
Step 8: Check option (D) 42.
This matches \(7 \times 6\) found in Step 4, so this option is correct.
Final Answer:
The two integers are 7 and 6, so their product is 42. \[ \boxed{P \times Q = 42} \]
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