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

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

Answer»

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.

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