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.