If P and Q are positive integers such that \(P^2 = Q^2 + 13\), find the value of the product \(PQ\).
Show 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.
Step 1: Algebraic Manipulation Rearrange the equation as a difference of squares: \[ P^2 - Q^2 = 13 \] Factor the left side: \[ (P - Q)(P + Q) = 13 \] Step 2: Integer Factorization Since \(P\) and \(Q\) are positive integers, both \((P-Q)\) and \((P+Q)\) must be integers. The number 13 is prime, so its only factors are 1 and 13. Since \(P\) and \(Q\) are positive, \((P+Q)\) must be greater than \((P-Q)\). Therefore: \[ P + Q = 13 \] \[ P - Q = 1 \] Step 3: Solving the System Add the two equations: \(2P = 14 \implies P = 7\). Subtract the equations: \(2Q = 12 \implies Q = 6\). The product \(PQ = 7 \times 6 = 42\).