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Edited: All equations in the post are assumed to have all real coefficients and are minimal polynomials.

While trying to ascertain if the Brioschi quintic $B(x)=x^5-10cx^3+45c^2x-c^2=0$ could ever have $3$ real roots, I was led to the question if one can use the discriminant $D$ to settle this.

For $B(x)$, it is given by $D = 5^5c^8(-1+1728c)^2$ and it seems for quintics, if $D>0$, then there are either $0$ or $4$ complex roots $C=a+bi$ with $b\neq0$. Hence the Brioschi (with real coefficients) can never have $3$ real roots.

For other degrees $n$, by observing the data in the Database of Number Fields, I was able to come up with the table below. The second and third columns give the number of complex roots $C=a+bi$.

$$\begin{array}{|c|c|c|} \hline \text{Degree}\;n&\text{If}\;D>0&\text{If}\;D<0\\ 2&0&2\\ 3&2&0\\ 4&{0,4}&{2}\\ 5&{0,4}&{2}\\ 6&2,6&0,4\\ 7&0,4&2,6\\ 8&{0,4,8}&{2,6}\\ 9&{0,4,8}&{2,6}\\ {10}&2,6,10&0,4,8\\ {11}&0,4,8&2,6,10\\ {12}&{0,4,8,12}&{2,6,10}\\ {13}&{0,4,8,12}&{2,6,10}\\ {14}&{0,4,8,12}&{2,6,10,14}\\ {15}&{0,4,8,12}&{2,6,10,14}\\ \hline \end{array}$$

Questions:

  1. Is the table true?
  2. How do we predict the second and third columns for much higher $n$? For example, for $n=163$, does the second column start as $0,4,8,12,\dots$ or $2,6,10,14,\dots$?
Community
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Tito Piezas III
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  • When $D<0$ and $n=5$ does the brown 2 refer to there being 2 complex roots? If so the remaining 3 would be real (but possibly not all distinct). [Maybe I'm missing something here...] I see you may be restricting to $D>0$ for some reason. – coffeemath Nov 30 '15 at 15:31
  • @coffeemath: Yes. That's what it means. Thus, if the table is true, if your quintic with real coefficients has $D<0$, then you can be sure it has only 2 complex roots (and 3 real roots). – Tito Piezas III Nov 30 '15 at 15:35
  • um excuse me but why aren't $2,7,11$ colored in red ? – mercio Nov 30 '15 at 16:38
  • @mercio: It was merely an error made in haste. – Tito Piezas III Nov 30 '15 at 16:47

2 Answers2

3

I believe that there is a mistake in your table.

Brill's theorem states that the sign of the discriminant of an algebraic number field is $(-1)^{r_2}$ where $r_2$ is the number of complex places. When we have a power basis for our number field, the minimal polynomial of the generator will have $2r_2$ complex roots. Thus the column for $D>0$ should only contain integers divisible by $4$, and when $D<0$ we have a number of complex roots $\equiv 2\pmod{4}$.

For a specific example, $$x^3-x^2-3x+1$$ has three real roots and a discriminant of 148.

Eric Naslund
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2

I don't think the table is correct. $f = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ has discriminant $1194393600 > 0$ and 0 complex roots.

It doesn't have a direct citation from a reputable source, but the Wikipedia page you linked says that in general for $D > 0$ there is an integer $0 \leq k \leq \frac{n}{4}$ such that there are $4 k$ complex roots, and for $D < 0$ there is an integer $0 \leq k' \leq \frac{n-2}{4}$ such that there are $4k' + 2$ complex roots, which doesn't mesh with your table.

Mark
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