Question:medium
\(P\) 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
The Correct Option is D
Solution and Explanation
Step 1: Try small values of Q directly:
Instead of factoring, we can test small positive integers for $Q$ and see when $Q^2 + 13$ becomes a perfect square.
Step 2: Test values one by one.
For $Q = 1$: $1 + 13 = 14$, not a perfect square. For $Q = 2$: $4 + 13 = 17$, not a perfect square. For $Q = 3$: $9 + 13 = 22$, not a perfect square. For $Q = 4$: $16 + 13 = 29$, not a perfect square. For $Q = 5$: $25 + 13 = 38$, not a perfect square. For $Q = 6$: $36 + 13 = 49$, and $49 = 7^2$, which is a perfect square.
Step 3: Confirm this is the only small solution.
As $Q$ grows larger, consecutive squares get further apart, so once $Q$ passes 6 the gap between $Q^2$ and the next square becomes bigger than 13, and no new solution can appear, so $Q = 6$ and $P = 7$ is the only pair that fits.
Step 4: Compute the product.
\[ P \times Q = 7 \times 6 = 42 \]
Step 5: Check option (A) 13.
13 is just the fixed number given in the problem, not a product of the two integers we found, so this option is wrong.
Step 6: Check option (B) 26.
26 does not equal $7 \times 6$, so this option is wrong.
Step 7: Check option (C) 39.
39 also does not equal $7 \times 6$, so this option is wrong.
Step 8: Check option (D) 42.
This equals $7 \times 6$ from Step 4, so this option is correct.
Final Answer:
Testing small values directly also confirms the product is 42. \[ \boxed{P \times Q = 42} \]
Instead of factoring, we can test small positive integers for $Q$ and see when $Q^2 + 13$ becomes a perfect square.
Step 2: Test values one by one.
For $Q = 1$: $1 + 13 = 14$, not a perfect square. For $Q = 2$: $4 + 13 = 17$, not a perfect square. For $Q = 3$: $9 + 13 = 22$, not a perfect square. For $Q = 4$: $16 + 13 = 29$, not a perfect square. For $Q = 5$: $25 + 13 = 38$, not a perfect square. For $Q = 6$: $36 + 13 = 49$, and $49 = 7^2$, which is a perfect square.
Step 3: Confirm this is the only small solution.
As $Q$ grows larger, consecutive squares get further apart, so once $Q$ passes 6 the gap between $Q^2$ and the next square becomes bigger than 13, and no new solution can appear, so $Q = 6$ and $P = 7$ is the only pair that fits.
Step 4: Compute the product.
\[ P \times Q = 7 \times 6 = 42 \]
Step 5: Check option (A) 13.
13 is just the fixed number given in the problem, not a product of the two integers we found, so this option is wrong.
Step 6: Check option (B) 26.
26 does not equal $7 \times 6$, so this option is wrong.
Step 7: Check option (C) 39.
39 also does not equal $7 \times 6$, so this option is wrong.
Step 8: Check option (D) 42.
This equals $7 \times 6$ from Step 4, so this option is correct.
Final Answer:
Testing small values directly also confirms the product is 42. \[ \boxed{P \times Q = 42} \]
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