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:
- Is the table true?
- 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 $2,6,10,14,\dots$?
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.