Question 4:

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

Answer:

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.

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