Solution:

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

Since the sum of the squares of the numbers = 365

x^{2} + (x + 1)^{2} = 365

x^{2} + [x^{2} + 2x + 1] = 365

x^{2} + x^{2} + 2x + 1 = 365

2x^{2} + a + 1 – 365 = 0

2x^{2} + 2x – 364 =0

x^{2} + x – 182 = 0

x^{2} + 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.**