For how many integer values of $a$ does the equation $x^2 + ax + 5a = 0$ have integer solutions for $x$?

To   have integer solutions.....the discriminant must  either possibly  = 0   or possibly be a  perfect square.....so

a^2 - 4(5a) = 0

a^2 - 20a  =  0

a ( a - 20) = 0

Setting each factor to 0 and solving for a   produces  a =  0   or  a = 20

Or

a^2 - 20a   = perfect square

a (a - 20)  = perfect square

This happens when   a = 20  , a = 0 , a = -16  or a  =  36

So....4 values for a  produce integer solutions