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Question
Find three consecutive odd integers, the sum of whose squares is 83.
Solution
Let the three numbers be x, x + 2, x + 4
According to statements,
(x)2 + (x + 2)2 + (x + 4)2 = 83
⇒ x2 + x2 + 4x + 4 + x2 + 8x + 16 = 83
⇒ 3x2 + 12x + 20 = 83
⇒ 3x2 + 12x + 20 - 83 = 0
⇒ 3x2 + 12x - 63 = 0
⇒ x2 + 4x - 21 = 0
⇒ x2 + 7x - 3x - 21 = 0
⇒ x(x + 7) -3(x + 7) = 0
⇒ (x - 3)(x + 7) = 0
Either x - 3 = 0,
then x = 3
or
x + 7 = 0,
then x = -7
∴ Number will be 3, 3 + 2, 3 + 4 = 3, 5, 7
or
Numbers will be -7, -7 + 2, -7 + 4 = -7, -5. -3.
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