The sum of the squares of two consecutive positive integers is 365. Find the integers.
Let the required two consecutive positive integers be x and x + 1
According to given condition,
x2 + (x + 1)2 = 365
x2 + x2 + 2x + 1 = 365 using (a + b)2 = a2 + 2ab + b2
2x2 + 2x – 364 = 0
x2 + x – 182 = 0
Using the splitting middle term – the middle term of the general equation 
is divided in two such values that:
Product = a.c
For the given equation a = 1 b = 1 c = – 182
= 1. – 182 = – 182
And either of their sum or difference = b
= 1
Thus the two terms are 14 and – 13
Difference = 14 – 13 = 1
Product = 14. – 13 = – 182
x2 + x – 182 = 0
x2 + 14x – 13x – 182 = 0
x(x + 14) – 13(x + 14) = 0
(x + 14) (x – 13) = 0
(x + 14) = 0 or (x – 13) = 0
x = – 14 or x = 13
x = 13 (x is a positive integer)
x + 1 = 13 + 1 = 14
Thus the required two consecutive positive integers are 13, 14