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

Solution:

Let x and x + 1 be the two consecutive positive integers.

Therefore, as per the given question,

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 + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

x = – 14 or x = 13

Since the integers are positive, x can be 13 only.

x + 1 = 13 + 1 = 14

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