Question

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

Answer 1

let the two numbers be a , a+1
given,
(a)²+(a+1)²= 365
a²+a²+2a+1 = 365
⇒2a²+2a+1 = 365
⇒2a²+2a-364= 0
⇒a²+a-182 = 0
⇒(a-13)x(a+14) = 0
⇒a = 13,-14
as they said positive integers
a=-14 is not possible
⇒a = 13

the two consecutive positive integers are 13,14

Answer 2

so the two numbers be x , x+1
so given

x² + (x+1)² = 365
x² + x² + 1 + 2x = 365
2x² + 2x - 364 = 0
x² + x - 182 = 0
x² + 14x - 13x - 182 = 0
x(x + 14) - 13(x + 14) = 0
(x - 13)(x + 14) = 0

so x = 13 
 so the 

consequetive numbers are 13,14