If the squares of three consecutive positive integers add up to 110, what are the three integers?

We denote the three consecutive integers as x, x+1 and x+2. Now
we are given that the sum of the squares of the three numbers is equal to 110. Therefore x^2 +
(x+1) ^2 +(x+2) ^2 = 110

=> x^2 + x^2 + 2x + 1 + x^2 + 4x + 4
= 110

=> 3x^2 + 6x + 5 =
110

=> 3x^2 + 6x – 105 =
0

=> x^2 + 2x – 35 = 0

=>
x^2 + 7x – 5x – 35 =0

=> x(x + 7) – 5(x + 7)
=0

=> (x – 5) (x +7)
=0

Therefore x can be 5 or -7.

So the
integers can be (5, 6, 7) or (-7, -6,-5)

The three
consecutive integers, the square of which adds up to 110 are (5, 6, 7) or (-7,
-6,-5).