Question 191906
let x, (x+2) & (x+4) be the three consecutive even integers.
(x+4)^2=(x+2)^2+100
x^2+8x+16=x^2+4x+4+100
x^2-x^2+8x-4x=100+4-16
4x=88
x=88/4
x=22 ans. for the first integer.
22+2=24 for the middle integer.
22+4=26 for the third integer.
Proof:
26^2=24^2+100
676=576+100
676=676