Answer :

**To Find:** Consecutive integers sum of whose square is 365

Consecutive integers mean that the difference between the integers is of 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

Now to factorize the above quadratic equation, we need to chose numbers such that their product is 182 and difference is 1

⇒ 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 question ask for positive integers, *x* can only be 13.

∴ *x* + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

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