Solution:
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.