Question

Find two consecutive natural numbers whose squares have the sum 221.

Answer 1

Let the number be x, x + 1
Then x² + ( x + 1)² = 221
⇒ x² + x² + 1 + 2x - 221 = 0
⇒  2x² + 2x - 220 = 0
⇒  x² + x - 110 = 0
⇒ x² + 11x - 10x - 110 = 0
⇒  x(x + 11) - 10(x + 11) = 0
⇒  (x = 11) (x - 10) = 0
⇒  x = -11 or x = 10
But x = -11 is rejected     ...[∵ It cannot been as its is a natural number]
∴ x = 10
Hence, required numbers are 10, 10 + 1.
i.e., 10 and 11.