Find two consecutive positive integers, sum of whose squares is 365.
Let the consecutive positive integers be x and x + 1.
Therefore, x2 + (x + 1)2 = 365
⇒ x2 + x2 + 1 + 2x = 365
⇒ 2x2 + 2x - 364 = 0
⇒ x2 + x - 182 = 0
⇒ x2 + 14x - 13x - 182 = 0
⇒ x(x + 14) -13(x + 14) = 0
⇒ (x + 14)(x - 13) = 0
Either x + 14 = 0 or x - 13 = 0,
⇒ 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.
