Requirements for an integer root of cubic equation

Question

If we have the quadratic equation $$a x^2 + b x + c = 0$$ with $a,b,c$ integers, then a requirement for $x$ to have an integer solution is for $b^2 - 4ac$ to be a square integer. This condition is necessary, but not sufficient. However it is simple enough to sometimes be useful when solving Diophantine equations.

Furthermore, note that this does not come from Vieta's formulas. While those are useful for other purposes, they do not yield a restrictive form in the integers like the condition on $b^2 - 4ac$ extracted from the general solution.

So with that introduction out of the way:
I would like to know if there are similar conditions for the cubic equation $$a x^3 + b x^2 + c x + d = 0.$$

In the quadratic case, if one root is integer, the other is at least rational. But in the cubic case, one root could be integer, with the others irrational or imaginary. So it looks like it would be harder to extract such a condition from the general solution (vanderbilt.edu, wikipedia). In particular, just like in casus irreducibilis where imaginary values will invariably show up during calculation of the roots even when all the roots are real, it seems inevitable that we could have irrationals like $\sqrt{n}$ show up only to cancel later in the calculation of an integer root.

So how can we extract from the general cubic solution some useful conditions on $a,b,c,d$ for an integer solution? Particularly nice, in analogy to the quadratic case, is if there is some term that is required to be a perfect cube.

Answer 1

As mentioned by stressed out, the rational root theorem is useful here. As stated, consider the general cubic equation

$$ax^3 + bx^2 + cx + d = 0 \tag{1}\label{eq1}$$

Also, have $d \neq 0$ since, otherwise, $x = 0$ is a root and \eqref{eq1} then reduces to a quadratic which you already know how to handle. As the Rational root theorem says, all rational roots of the form $x = p/q$, where $p$ and $q$ are in lowest terms, require that $p \mid d$ and $q \mid a$. The integral solutions, of course, would be a subset of this where $q = 1$. As such, just consider $p$ being a root. Also, $p \mid d$ means there is some integer $e \neq 0$ such that

$$d = pe \tag{2}\label{eq2}$$

Substituting $x = p$ and \eqref{eq2} into \eqref{eq1}, then dividing by $p$, gives

$$ap^2 + bp + \left(c + e\right) = 0 \tag{3}\label{eq3}$$

Treating $e$ as a constant for now, this is a quadratic equation in $p$. Thus, using the quadratic formula means there is only possibly such an integer $p$ if

$$b^2 - 4a\left(c + e\right) = f^2 \tag{4}\label{eq4}$$

is a perfect square, i.e., $f$ is an integer. As such, you just need to check the various divisors (both positive & negative) $e$ of $d$ in \eqref{eq4}.

I don't believe this is quite what you're looking for, but I'm not sure if you will be able to find anything much better, especially just one specific equation to check. This is because any such alternative formulation would need to basically be equivalent to \eqref{eq4} which generally has several values to check. Nonetheless, I could definitely be wrong, with somebody providing a simpler formulation for you to use.