Question

Find two consecutive positive integers, sum of whose squares is 613.

Answer 1

Let a positive integer be x. 

Then the second integer = x + 1 

Sum of the squares of the above integers = x² + (x + 1)² 

= x² + x² + 2x + 1 

= 2x² + 2x + 1 

By problem 2x² + 2x + 1 = 613 

⇒ 2x² + 2x – 612 = 0

⇒ x² + x – 306 = 0 

⇒ x² + 18x – 17x – 306 = 0 

⇒ x(x + 18) – 17(x + 18) = 0 

⇒ (x – 17) (x + 18) = 0 

⇒ x – 17 = 0 (or) x + 18 = 0 

⇒ x = 17 (or) -18,

we do not consider -18 

Then the numbers are (17, 17 + 1) 

i.e., 17, 18 are the required two consecutive positive integers.