Question

Find three consecutive odd integers ,the sum of whose squares is 83

Answer 1

Let smallest odd number be x
Therefore,next odd no.= (x+2)
the odd no. after that = (x+2)+2 = (x+4)

As per question,
=> x^2+(x+2)^2+(x+4)^2 = 83
=> x^2+(x^2+4x+4)+(x^2+8x+16) = 83
=> 3x^2+12x+20 = 83
=> 3x^2+12x-63 = 0
Dividing equation by 3
=> x^2+4x-21 = 0
=> x^2+7x-3x-21 = 0
=> x(x+7)-3(x+7) = 0
=> (x-3)(x+7) = 0
Therefore,
=> (x-3) = 0 or (x+7) = 0
=> x = 3 or x = -7

If x = 3, then remaining numbers are
=> (3+2) = 5
and
=> (3+4) = 7

If x = -7, then remaining numbers are
=> (-7+2) = -5
and
=> (-7+4) = -3

Therefore, the numbers can be 3, 5 and 7 OR -7, -5 and -3.