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.
 A: The discriminant of $x^3-27x+k$ is $-27(k^2-54^2)$. It follows that
$$ x^3-27 x = -k $$
has $\geq 2$ distinct real solutions for each $k$ in the range $[-54,54]$.

By Vieta's formulas, if $x^3-27x+k$ has two distinct integer roots $a,b$ then $-(a+b)$ is also an integer root. Additionally we have $ab(a+b)=k$ and $a^2+ab+b^2 = 27$. $a^2+ab+b^2$ is a quadratic form with a negative discriminant, hence $a^2+ab+b^2=27$ has a finite number of integer solutions. They are simple to find by factoring both sides over $\mathbb{Z}[\omega]$ (which is a UFD):
$$(a,b)\in\{(-6,3),(-3,-3),(-3,6),(3,-6),(3,3),(6,-3)\}$$
hence there are just $\color{red}{2}$ values of $k$ doing the job, namely $\pm 54$.
A: 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 $-27$: $x^2-27x\pm 54$.
