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

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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).**

Let the three consecutive numbers be x-1, x and x+1 .

So the sum of the squares of the consecutive numbers = (x-1)^2+x^2+(x+1)^2 = (x^2-2x+1)+x^2+(x^2+2x+1).= 3x^2+2. But the actual sum is given to be 110.

Therefore 3x^2+2 = 110.

Therefore 3x^2 = 110 -2.

So 3x^2 = 108.

We divide both sides of the equation by 3.

x^2 = 108/3 = 36.

x^2= 36.

x = 6.

Therefore the consecutive numbers : 6-1 , 6 and 6 +1. Or they are 5, 6 and 7.

Tally: sum the squares of the numbers: 5^+6^2+7^2 = 25+36+49 = 110.