All zeros of a polynomial of degree 3 are positive integers

Question

Let us assume that we have a polynomial of the form $$x^3+x^2a+xb+c$$ Which conditions are necessary (or sufficient) for $a$, $b$, and $c$ to guarantee that all three roots of the function (counting multiplicity) are positive integers?

score 1 · Accepted Answer · answered Nov 26 '19 at 16:02

Well, you need the coefficients to be integers and the signs in the polynomial to alternate (these are necessary conditions), so lets rewrite the polynomial as $x^3-a x^2 + b x - c$ with integers $a, b, c > 0$.

Now, we need some positive divisor $d | c$, to be a root, so $d^3-ad^2+bd-c=0 \implies d^2-ad+(b-c/d) = 0$ is a quadratic (in $d$ now) which needs two positive integer roots. Necessarily $b>c/d$ and then all we need is for $\Delta=a^2-4b+4c/d$ to be a perfect square.

Putting all of that together, the necessary and sufficient conditions are, the cubic is of form $x^3-ax^2+bx-c$ with integers $a, b, c > 0$ and $\exists d \in \mathbb Z$ s.t. $d|c$, $bd>c$ and $a^2-4b+4c/d$ is a perfect square. We of course only need to test for divisors up to $d \leqslant \sqrt[3]c$

———
P.S. It may be simpler to check the divisors of $c$ using the factor theorem and find the roots or conclude otherwise; rather than test them for the perfect square condition.

On the signs alternating, search for Descartes’ rule of signs. Alternately, study the coefficients of $(x-\alpha)(x-\beta)$ and $(x-\alpha)(x-\beta)(x-\gamma)$ by simply expanding, when $\alpha, \beta, \gamma$ are positives. — Macavity, Nov 26 '19 at 17:05

José Carlos Santos · Answer 2 · 2019-11-26T14:50:43.173

0

Since it is a monic polynomial, it is necessary that $a,b,c\in\mathbb Z$ and, since the roots are positive, that $a,c<0<b$. And it follows from the rational root theorem that it is sufficient that it has $3$ roots (counting them with their multiplicities) among the positive divisors of $c$.

edited Nov 26 '19 at 14:50

answered Nov 26 '19 at 14:15

José Carlos Santos

427,504

are positive integers... – Alexey Burdin Nov 26 '19 at 14:36
Thanks. I had missed that hypothesis. I've edited my answer. – José Carlos Santos Nov 26 '19 at 14:40
2

Your second "sufficient" condition is just saying that if the roots are all integers, they must be among the positive divisors of $c$. I suspect the OP is looking for a statement more along the lines of "If $a$, $b$, $c$ satisfy the following conditions, then the roots are all integers." – Michael Seifert Nov 26 '19 at 14:55
@MichaelSeifert Me too, but this was the best I could do. – José Carlos Santos Nov 26 '19 at 14:59
@MichealSeifert is right, but thanks for your answer anyways – blausup Nov 26 '19 at 15:18

score 0 · Answer 3 · answered Nov 26 '19 at 15:24

As pointed out by Andrew Chin and José Carlos Santos a necessary condition is that

(1) $a,b,c$ are integers such that $a,c < 0$ and $b > 0$.

We can say a little more. Setting $x = y - \frac{a}{3}$ we get the equation $$(*) \quad y^3 + ry = s$$ where $r = b -\frac{a^2}{3}$ and $s = -\frac{2a^3}{27}+\frac{ab}{3}-c$. Equation $(*)$ can be solved by the Cardano formula. See for example my answer to Is there really analytic solution to cubic equation?

In order that $(*)$ has three real solutions it is necessary and sufficient that $$R = \frac{s^2}{4} + \frac{r^3}{27} \le 0 .$$ Inserting gives

(2) $4b^3 -a^2b^2 +27c^2 +4a^3c -18abc \le 0$

Unfortunately this does not give any information whether the three real roots are positive or integers.

All zeros of a polynomial of degree 3 are positive integers

3 Answers3