jmc

Suppose the roots of the quadratic are given by $m$ and $n$. Note that $$(x-m)(x-n)=x^2-(m+n)x+mn=x^2+ax+5a,$$ and setting coefficients equal, it follows that $$\begin{array}{rl}m+n& =-a\\ mn& =5a\end{array}$$ (This also follows directly from Vieta's formulas.) Notice that the $a$ can be canceled by either dividing or noting that $$0=5a+5\cdot (-a)=mn+5(m+n).$$

Simon's Favorite Factoring Trick can now be applied: $$mn+5m+5n+25=(m+5)(n+5)=25.$$ It follows that $m+5$ and $n+5$ are divisors of $25$, whose pairs of divisors are given by $\pm \{(1,25),(5,5),(25,1)\}$. Solving, we see that $(m,n)$ is in the set $$\{(-4,20),(0,0),(20,-4),(-6,-30),(-10,-10),(-30,-6)\}.$$ However, the two pairs of symmetric solutions yield redundant values for $a$, so it follows that the answer is $\boxed{4}$.