The number of integers $k$ for which the equation $x^3-27x+k$ has at least two distinct integer roots is? The number of integers $k$ for which the equation $x^3-27x+k$ has at least two distinct integer roots is?
My try:
Since imaginary numbers always form roots in pairs the question is the same as having two integer roots and one other real root.
If the roots are $A$,$B$ and $C$ of which $A$ and $B$ are two distinct integers and $C$ is a real number then we can write 
1.
$$-k=ABC$$
2.
$$-27=AB+BC+CA$$
3.
$$A+B+C=0$$
Equation 3. Implies that even $C$ has to be an integer.
Now I don't know how I can find the number of values of k for which all roots are integers.

Substituting $c = -\frac{k}{ab}$ into the third equation gives
$$a+b-\frac{k}{ab}=0\quad\Rightarrow\quad k = ab(a+b).$$
Using that in the second equation and simplifying gives
$$(a+b)^2 = ab+27,\text{ or }a^2+ab+b^2=27.$$
Thus
$$b = \frac{1}{2}(-a\pm\sqrt{3(36-a^2)}\,).$$
Since $a$ and $b$ must both be integers, one quickly gets the pairs
$$(a,b) = (-6,3), (-3,-3), (-3,6), (3,-6), (3,3), (6,-3),$$
each of which gives a solution. However, only two of those solutions have the linear term equal to $-27$: $x^2-27x\pm 54$.
