imc

By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral. Because the zeros are integral, the discriminant of the function, $a^2-8a$, is a perfect square, say $k^2$. Then adding 16 to both sides and completing the square yields $$(a-4{)}^2=k^2+16.$$ Therefore $(a-4{)}^2-k^2=16$ and $$((a-4)-k)((a-4)+k)=16.$$ Let $(a-4)-k=u$ and $(a-4)+k=v$; then, $a-4=\frac{u+v}{2}$ and so $a=\frac{u+v}{2}+4$. Listing all possible $(u,v)$ pairs (not counting transpositions because this does not affect ($u+v$), $(2,8),(4,4),(-2,-8),(-4,-4)$, yields $a=9,8,-1,0$. These $a$ sum to $16$, so our answer is $\boxed{16}$.