The difference between the squares of two consecutive even integers is always divisible by:

Let the two consecutive even integers be $2n$ and $2n+2$.
The difference between their squares is:

$$(2n+2{)}^2-(2n{)}^2=(4n^2+8n+4)-4n^2=8n+4=4(2n+1)$$

Since $4(2n+1)$ is explicitly a multiple of $4$, the difference is always completely divisible by $4$. (Note: It is also divisible by 4 but not always 8 because $2n+1$ is always odd).