Question

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

Answer 1

Let the two consecutive positive integers be x and (x + 1).

                   Since the sum of the squares of the numbers = 365

                   x² + (x + 1)² = 365

                   x² + [x² + 2x + 1] = 365

                   x² + x² + 2x + 1 = 365

                   2x² + a + 1 – 365 = 0

                   2x² + 2x – 364 =0

                   x² + x – 182 = 0

                   x² + 14x – 13x – 182 = 0                   ∵ +14 –13 = 1 and 14 × (–13) = ‐ 182

                   x(x + 14) –13 (x + 14) = 0

                   (x +14) (x – 13) = 0

                   Either x + 14 = 0 ⇒ x = – 14

                   or x – 13 = 0 = x = 13

                   Since x has to be a positive integer

                   ∴ x = 13

                   ⇒ x + 1 = 13 + 1 =1 4

                   Thus, the required consecutive positive integers are 13 and 14.