Find two consecutive positive integers, sum of whose squares is 365.
Let the consecutive positive integers be x and x + 1.
Given that 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
Either x + 14 = 0 or x − 13 = 0, i.e., x = −14 or x = 13
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.