Find three consecutive even integers such that the square of the third is 4 more than the square of the second.

Answer 1

1st is x

2nd is x+2

3rd is x+4

(x+4)^2 = (x+2)^2 +4

x^2 + 8x + 16 = x^2 + 4x + 4 +4

x^2 + 8x + 16 = x^2 + 4x + 8

8x + 16 = 4x + 8

4x + 8 = 0

x = -8/4 = -2

They are:

-2, 0, 2