Question

Find two consecutive positive integers, sum of whose squares is 365.

Answer 1

Let the consecutive positive integers be x and x + 1. 

Given that \(x^2 + (x+1)^2 = 365\)

⇒ \(x^2 + x^2 +1 +2x =365\)
⇒ \(2x^2 +2x -364=0\)
⇒ \(x^2 +x -182=0\)
⇒ \(x^2 +14x -13x -182 =0\)
⇒ \(x(x+14) -13(x+14) =0\)
⇒ \((x+14)(x+13) =0\)

Either x + 14 = 0 or x − 13 = 0, 
i.e., x = −14 or x = 13

Since the integers are positive, x can only be 13. 
∴ x + 1 = 13 + 1 = 14 

Therefore, two consecutive positive integers will be 13 and 14.