problem: For how many integer values of $a$ does the equation $x^2 + ax + 5a = 0$ have integer solutions for $x$?
solution: 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{align*}
  m + n &= -a \\
  mn &= 5a
  \end{align*} (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 $\boxed{4}$.