Any even integer takes the form [tex]2n[/tex], where [tex]n[/tex] is any integer. Even integers occur every two integers following some even integer. This means the next even integer will take the form [tex]2n+2[/tex].
So you want to find [tex]2n(2n+2)=4n^2+4n[/tex].
If you're told that the two integers add to some number, call it [tex]K[/tex], then we also know
[tex]2n+(2n+2)=4n+2=K[/tex]
and from that we can solve for [tex]n[/tex], which is exactly what we need to know in order to get a concrete value for the product above. In particular,
[tex]4n+2=K\implies n=\dfrac{K-2}4[/tex]
and so
[tex]2n(2n+2)=4n^2+4n=(K-2)^2+(K-2)=K^2-3K+2[/tex]
