We all know that if the discriminant of a monic quadratic is a perfect square, then both of its roots will be integers. In my research, I'm interested in monic cubics, and I was wondering whether there is some condition on the discriminant of a cubic, such that all it's roots are integers? I'm guessing it's something along the lines of the discriminant is a perfect cube?

I have already researched this, trust me, but to no avail. Every page I found described the nature of the roots as vaguely "all real", "two complex, one real", etc. instead of whether or not they are integral, or even rational. This lead to me to believe that such a condition doesn't exist, but I wanted to make sure by posting a question here. If the condition exists, what is it, and how do you prove it?

My very first question on this site; thanks in advance!

  • $\begingroup$ Have you found a sufficient condition for this problem? $\endgroup$
    – nls
    Oct 9 '16 at 20:41

Let me start with the following suggestion. Suppose you conjecture that there should be some relationship between cubics with integral roots and the discriminant, but let's say that you are at a loss on how to theoretically find any relationship. Then the first thing any (aspiring) mathematician should do is compute some examples! I believe you when you say you tried searching the web for answers, but that is not how mathematicians do research when they get stuck --- rather, you should play around with simple examples. And if you had tried to compute the discriminant of the simplest possible cubic with distinct integer roots, then you would have found that the discriminant of

$$x^3 - x = (x-1)x(x+1)$$

was equal to $4$, which is not a cube.

More generally, there are two ways to think about the discriminant. One is some complicated formula you find in a book, the other (more naturally) is given as a formula in the roots, namely, the discriminant of

$$(x - \alpha)(x - \beta)(x - \gamma)$$

is $(\alpha - \beta)^2 (\alpha - \gamma)^2 (\beta - \gamma)^2$. From this, you can see that if all the roots are integers, then the discriminant will be a square. This is true for any degree. Since replacing $x$ by $x - \gamma$ doesn't change the discriminant, you can also restrict to polynomials of the form:

$$f(x) = x(x-a)(x-b),$$

which has discriminant

$$\Delta = a^2 b^2 (a-b)^2.$$

So certainly a necessary condition to have integral roots is that $\Delta$ be of this form. You might guess that being a square is enough to ensure integral roots. However, after playing around for a while, you might find the example of the irreducible polynomial

$$x^3 - x^2 - 2x + 1$$

with discriminant $49$. So being a square is not enough. On the other hand, there are no solutions to $7^2 = a^2 b^2 (a-b)^2$, because one of $a$, $b$ or $(a-b)$ would have to be $7$, and then the other two would have to be $\pm 1$ which is not possible.

If you take a first course in Galois theory, you will learn that the Galois closure of a polynomial $f(x)$ of degree $n$ has a Galois group $G$ which is a subgroup of the symmetric group $S_n$. The condition that the discriminant is a square is exactly the condition that $G$ is actually a subgroup of $A_n$. So, for quadratics, having square discriminant is the same as having trivial Galois group, and having square discriminant for a cubic is the same has having trivial Galois group or Galois group $\mathbf{Z}/3\mathbf{Z} = A_3$. (This is how I found the cubic above, it comes from the real subfield of $\mathbf{Q}(\zeta_7)$.)

But at least there is one characterization:

If $\Delta$ is square and $f(x)$ has at least one integer root, then it has three integer roots.

I'll let you prove that. Finally, here's an example to show that even being of the form $a^2 b^2(a-b)^2$ is not enough:

$$x^3 - 9 x^2 - x + 1 \ \text{is irreducible and has discriminant} \ 3136,$$ $$x(x-1)(x-8) \ \text{has discriminant} \ 3136.$$

  • $\begingroup$ Oh, and by the way, if a polynomial $f(x)$ with integral coefficients has rational roots, then all the roots are integral --- this is a special case of Gauss's Lemma, and is also easy to prove directly. $\endgroup$ Sep 28 '14 at 18:17
  • $\begingroup$ Thanks! I'm too short into my mathematical career to understand the Galois bit, but nice solution! $\endgroup$ Sep 28 '14 at 18:28

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