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


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