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

Dear Student

Let us say,
the two consecutive positive integers be x and x + 1.
Therefore, as per the given questions, x^2 + (x + 1)^2 = 365
⇒ x^2 + x^2 + 1 + 2x = 365
⇒ 2x^2 + 2x - 364 = 0
⇒ x^2 + x - 182 = 0
⇒ x^2 + 14x - 13x - 182 = 0
⇒ x(x + 14) -13(x + 14) = 0
⇒ (x + 14)(x - 13) = 0
Thus, either, x + 14 = 0 or x - 13 = 0,
⇒ x = - 14 or x = 13
since, the integers are positive, so x can be 13,
x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14


Thanks