Question

Find two consecutive odd integers such that the sum of their squares is 394.

Answer 1

Let first odd integer = 2x + 1
Then second odd integer = 2x + 3
According to the condition,
(2x + 1)² + (2x + 3)² = 394
⇒ 4x² + 4x + 1 + 4x² + 12x + 9 = 394
⇒ 8x² + 16x - 394 + 10 = 0
⇒ 8x² + 16x - 384 = 0
⇒ x² + 2x - 48 = 0     ...(Dividing by 8)
⇒ x² + 8x - 6x - 48 = 0
⇒ x(x + 8) -6(x + 8) = 0
⇒ (x + 8)(x - 6) = 0
EIther x + 8 = 0,
then x = -8
or
x - 6 = 0,
then x = 6
(i) If x = -8, then first odd integer = 2x + 1
= 2 x (-8) + 1
= -16 + 1
= -15
(ii) If x = 6, then first odd integer = 2x + 1
= 2 x 6 + 1 = 13
and second integer = 13 + 2 = 15
∴ Required integers are -15, -13, or 13, 15.