Construct a degree 3 monic polynomial with integer coefficients that has 3 irrational roots.

Question

The polynomial $x^3 - 3x + 1$ is monic, degree $3$, has integer coefficients and all its roots are irrational. I found this polynomial using Mathematica to generate random polynomials and then selecting the ones that have $3$ real roots. My code is not efficient and I am unable to generate any significant number of examples.

Is there a way to construct such a polynomial? If we restrict the range of the integer coefficients $(-10,10)$ is there a way to count exactly how many such polynomials exist?

Answer 1

The discriminant of a monic, cubic polynomial $$p(x) := x^3 + b x^2 + c x + d$$ is the invariant $$\Delta := -4 b^3 d+b^2 c^2+18 bc d-4 c^3-27 d^2;$$ it has the useful property that $p$ has three distinct, real roots iff $\Delta > 0$. (If $p$ has a repeated real root, then $\Delta = 0$, but in this case the nonrepeated root is always rational.)

On the other hand, by the Rational Root Theorem, if a monic polynomial $x^n + \cdots + d$ has a rational root $r$, then $r$ is an integer that divides $d$. These facts together suggest a way of generating examples:

1. Pick a triple $(b, c, d)$ of integers.
2. Compute $\Delta$; if $\Delta \leq 0$, $p$ does not have three real roots, so start over and pick a new triple. Otherwise, $p$ has three real roots.
3. For each of the factors $s$ of $d$. Computing $p(\pm s)$. If any of these is values is zero, then $p$ has a rational root, so start over and pick a new triple. Otherwise, if none of these is zero, none of the roots of $p$ are rational, that is, $p$ satisfies the condition.

A quick Maple script shows that $2922$ ($31.2\%$) of the $21^3 = 9261$ monic, cubic polynomials with integer coefficients in $-10, \ldots, 10$ satisfy the condition, so the above procedure is efficient in the sense that in practice, one needn't try too many triples $(b, c, d)$ to produce examples.

Answer 2

If $$ x^3+ax^2+bx+c$$ (with integers $a,b,c$) has a rational root, then it is in fact an integer root and is a (positive or negative) divisor of $c$ (rational root theorem). In particular, if you let $c=\pm1$, you need only ensure that $\pm1$ are no roots, i.e., that $a+b+c+1\ne 0$ and $-1+a-b+c\ne 0$. If $c$ is $\pm$ a prime there are only a few more conditions and if $c$ is composite, the situation is still not difficult to handle as long as we know the factorization of $c$.

However, in order to exclude non-real roots, we have to do one additional check: The cubic above will have a pair of conjugate complex roots iff its discriminant $$\Delta=a^2b^2-4b^3-4a^3c-27c^2+18abc$$ is negative. Using these two tests it is not too hard to determine whether a given cubic has only real-irrational roots.