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Consider the following problem:

Let $p\in\mathbb{Z}[x]$ be a polynomial with integer coefficient. Suppose that the leading coefficient is 1, all roots are real and in $(0, 3)$. Find all possible roots of $p$.

The roots 1 and 2 are abvious, but there are more possible roots: the roots of $x^2-3x+1$, $$\frac{3+\sqrt{5}}{2}\quad\text{ and }\quad\frac{3-\sqrt{5}}{2}.$$

Other possible roots I didn't find, but what is interesting (at least for me) I notice that the above non-integer roots have the form $$\frac{3+\sqrt{5}}{2} = 1+\phi\quad\quad\frac{3-\sqrt{5}}{2} = 2-\phi,$$ where $$\phi = \frac{\sqrt{5}+1}{2}$$ is the golden ratio constant.

  1. Is the set of all possible roots $\{1, 1+\phi, 2-\phi, 2\}$ ?
  2. (if the answer of question 1 is positive) Why the golden ratio constant appears in this problem ?
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    $\begingroup$ I'm slightly confused. So is your question to find all real algebraic integers between 0 an 3? Yes $ \phi$ is one of them, but not all of them have anything to do with $\phi$ $\endgroup$ – Mathmo123 Aug 5 '14 at 13:26
  • $\begingroup$ @Mathmo123: Not just real algebraic number, for example, $\phi$ is not one of them, because it is the root of $p(x)=x^2-x-1$, but the other root of $p$ is $$\frac{1-\sqrt{5}}{2}\not\in (0,3)$$ $\endgroup$ – Mher Aug 5 '14 at 13:49
  • $\begingroup$ I don't have time to chase this up right now, but I think the question is related to the concept of "transfinite diameter" of an interval. $\endgroup$ – Gerry Myerson Aug 6 '14 at 7:37
  • $\begingroup$ @Mher: Alternatively,$$\frac{3+\sqrt{5}}{2}=\phi^2,\;\;\frac{3-\sqrt{5}}{2}=1/\phi^2$$ $\endgroup$ – Tito Piezas III Jun 18 '15 at 6:19
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This answer is only for degree $2$.

There are only four pairs $(a,b)=(-2,1),(-3,1),(-3,2),(-4,4).$ (Maybe you want to eliminate $(-2,1),(-4,4)$, though.)

If $x^2+ax+b=0\ (a,b\in\mathbb Z)$, then we have $$0\lt \frac{-a-\sqrt{a^2-4b}}{2}\lt 3,\ \ 0\lt\frac{-a+\sqrt{a^2-4b}}{2}\lt 3$$ $$\iff 0\lt -a-\sqrt{a^2-4b}\lt 6,\ \ 0\lt -a+\sqrt{a^2-4b}\lt 6\tag1$$ Since we have $0\lt -2a\lt 12\iff -6\lt a\lt 0$, $$(1)\iff a\lt -\sqrt{a^2-4b}\lt a+6,\ \ a\lt\sqrt{a^2-4b}\lt a+6$$ $$\iff a\lt -\sqrt{a^2-4b},\ \ \sqrt{a^2-4b}\lt a+6$$ $$\iff -a\gt \sqrt{a^2-4b},\ \ \sqrt{a^2-4b}\lt a+6$$ $$\iff (-a)^2\gt a^2-4b,\ \ a^2-4b\lt (a+6)^2\iff b\gt 0,\ \ b\gt -3a-9.$$

With $a^2-4b\ge 0$, we have $$-6\lt a\lt 0,\ \ b\gt 0,\ \ b\gt -3a-9,\ \ b\le \frac{a^2}{4}.$$

Considering these conditions on $ab$ plane will give you all possible pairs $(a,b)$, which are the followings :

$$(a,b)=(-2,1),(-3,1),(-3,2),(-4,4)$$

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  • $\begingroup$ OK, greate, but it is not supposed that $\deg p=2$. $\endgroup$ – Mher Aug 5 '14 at 13:10
  • $\begingroup$ @Mher: Oh, my bad. I got it. $\endgroup$ – mathlove Aug 5 '14 at 14:25
  • $\begingroup$ What about $(-2,1)$? Like $(-4,4)$ was the the repeated root at $2$ right? $\endgroup$ – snulty Aug 6 '14 at 7:21
  • $\begingroup$ @snulty: Oh, you are right. Thanks. $\endgroup$ – mathlove Aug 6 '14 at 8:23
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Polya proved that any interval of length less than 4 contains only finitely many sets of conjugate algebraic integers (in the language of the original question, only finitely many polynomials with integer coefficients, leading coefficient 1, all roots real and in any given interval of length less than 4). I'm not sure where Polya proved this. It might be in his paper Ueber Ganzwertige Ganze Funktionen, Rend Circ Mat Palermo 40 (1915) 1-16.

Raphael M Robinson, Algebraic equations with span less than 4, Math Comp 18 (1964) 547-559, MR 0169374, 29 #6624 (I believe the paper is freely available on the American Math Society website) found all the irreducible polynomials of degree between 2 and 6, integer coefficients, leading coefficient 1, all roots real, greatest root minus smallest ("span") less than 4, and also several such polynomials of degrees 7 and 8.

It was later proved by Capparelli et al that Robinson's list was complete through degree 8.

The only ones with span under 3 were the ones OP found with $\sqrt5$, and the polynomial $x^2-2$. But there's no way to translate $x^2-2$ so its roots are between $0$ and $3$ and stay algebraic integers.

The ones with higher degree all have span close to 4, so I'm convinced that OP has found all the ones with roots between $0$ and $3$, even though I haven't been able to find a paper that says exactly that.

It might be worth having a look at Capparelli et al, On the span of polynomials with integer coefficients, Math Comp 79 (2010) #270 967-981, MR 2600551, 2011b:12001, where Robinson's computations are extended to degree 14, just to see whether it gives any history, or cites Polya, or mentions span 3.

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