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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?

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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.

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    $\begingroup$ -1 No. It does not have integer coefficients (in general). $\endgroup$ – almagest Apr 16 '16 at 14:21
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    $\begingroup$ 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
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    $\begingroup$ Ok, but that misses most of them. $\endgroup$ – almagest Apr 16 '16 at 14:22
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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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

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