Question

If P and Q are positive integers such that \(P^2 = Q^2 + 13\), find the value of the product \(PQ\).

• A. 13
• B. 42
• C. 49
• D. 36

Hint:

When you encounter an equation of the form \(x^2 - y^2 = p\), where \(p\) is a prime number, immediately factor it as \((x-y)(x+y) = p \times 1\). This quickly sets up a system of two linear equations: \(x+y=p\) and \(x-y=1\), which is very easy to solve.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given an equation relating the squares of two positive integers, P and Q. Our goal is to find their values and then calculate their product, \(PQ\).
Step 2: Key Formula or Approach:
The key to solving this problem is to rearrange the equation and use the algebraic identity for the difference of squares: \[ a^2 - b^2 = (a-b)(a+b) \] We will also use the fact that 13 is a prime number.
Step 3: Detailed Explanation:
1. Rearrange the given equation: \[ P^2 = Q^2 + 13 \] \[ P^2 - Q^2 = 13 \] 2. Apply the difference of squares formula: \[ (P - Q)(P + Q) = 13 \] 3. Analyze the factors: Since P and Q are positive integers, \((P - Q)\) and \((P + Q)\) must also be integers. Furthermore, since \(P>0\) and \(Q>0\), their sum \((P + Q)\) must be a positive integer. Because their product is 13 (a positive number), the term \((P - Q)\) must also be a positive integer. The number 13 is a prime number, which means its only positive integer factors are 1 and 13. Also, since Q is positive, \((P+Q)>(P-Q)\). Therefore, we must have: \[ P + Q = 13 \] \[ P - Q = 1 \] 4. Solve the system of linear equations: We now have a simple system of two equations. We can solve it by adding the two equations together: \[ (P + Q) + (P - Q) = 13 + 1 \] \[ 2P = 14 \] \[ P = 7 \] Now, substitute the value of P back into the first equation: \[ 7 + Q = 13 \] \[ Q = 13 - 7 \] \[ Q = 6 \] 5. Calculate the product PQ: \[ PQ = 7 \times 6 = 42 \] Step 4: Final Answer:
The values of the integers are P = 7 and Q = 6. Their product, PQ, is 42.