jmc

Suppose the roots of the quadratic are given by $m$ and $n$ with $m\le n$. Note that $$(x-m)(x-n)=x^2-(m+n)x+mn=x^2+ax+8a,$$ and setting coefficients equal, it follows that $$\begin{array}{rl}m+n& =-a\\ mn& =8a\end{array}$$ (This also follows directly from Vieta's formulas.) Adding $8$ times the first equation to the second gives us that $$8(m+n)+mn=0$$ Simon's Favorite Factoring Trick can now be applied by adding $64$ to both sides: $$mn+8m+8n+64=(m+8)(n+8)=64.$$ It follows that $m+8$ and $n+8$ are divisors of $64$, whose pairs of divisors are given by $\pm \{(1,64),(2,32),(4,16)$ and $(8,8)\}$. Solving, we see that $(m,n)$ must be among the pairs $$\begin{array}{rl}& (-7,56),(-6,24),(-4,8),(0,0),\\ & (-72,-9),(-40,-10),(-24,-12),(-16,-16).\end{array}$$ Since $a=-(m+n)$ and each of these pairs gives a distinct value of $m+n$, each of these $8$ pairs gives a distinct value of $a$, so our answer is $\boxed{8}$.