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?


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.

  • 1
    $\begingroup$ -1 No. It does not have integer coefficients (in general). $\endgroup$ – almagest Apr 16 '16 at 14:21
  • 1
    $\begingroup$ Of course, I meant to specify that $a, b, c$ are integers, not reals, which I've now done. $\endgroup$ – Travis Apr 16 '16 at 14:22
  • 1
    $\begingroup$ Ok, but that misses most of them. $\endgroup$ – almagest Apr 16 '16 at 14:22
  • 1
    $\begingroup$ Yes, of course, there's no claim that these are exhaustive, and indeed, the example in the question cannot be written this way; OP simply asked for a method of generating examples. $\endgroup$ – Travis Apr 16 '16 at 14:25
  • 1
    $\begingroup$ @Travis. OK Thanks. I agree with your argument and your computation of 2922. I also have the number of such cubics with coefficients in -n, ... , n as: 0, 6, 32, 104, 248, 484, 850, 1356, 2046, 2922 for n=1,..,10. $\endgroup$ – Geoffrey Critzer Apr 16 '16 at 16:04

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.

  • $\begingroup$ But he presumably wants (1) all roots real as well as (2) no roots rational. $\endgroup$ – almagest Apr 16 '16 at 14:30
  • $\begingroup$ @almagest: 2 is covered-the roots will be irrational. $\endgroup$ – Ross Millikan Apr 16 '16 at 14:31
  • $\begingroup$ @almagest Yes. I want a polynomial that has all 3 roots that are irrational. No nonreal complex roots are allowed. $\endgroup$ – Geoffrey Critzer Apr 16 '16 at 14:32
  • $\begingroup$ @GeoffreyCritzer Ah, ok. For me nonreal complex roots are also irrational (as they are not rational) ... - You'll need to use the discriminant as additional criterion then. $\endgroup$ – Hagen von Eitzen Apr 16 '16 at 14:33
  • $\begingroup$ @HagenvonEitzen yes. I am sorry for the confusion with my question. I should have stated it more precisely. I define irrational to be a subset of the reals. $\endgroup$ – Geoffrey Critzer Apr 16 '16 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.