Let $n \geq 3$ be a natural number. Define the set $X_n$ as the set of natural numbers that appear as the number of real roots an irreducible polynomial of degree $n$ over $\mathbb{Q}$ which is solvable by radicals can have.

Example: In case $n=p$ is a prime, we have $X_p= \{1,p \}$.

Is $X_n$ or the cardinality $|X_n|$ also known for other values?

It would be interesting to see the beginning of the sequence $a_n=|X_n|$ for small values of $n$, maybe it appears in the oeis.

  • $\begingroup$ what (trivial or nontrivial) things do you know about $X_n$ in general? $\endgroup$ – Arno Fehm May 10 '20 at 20:20
  • $\begingroup$ @ArnoFehm I dont really know much about $X_n$ except for the stated result for prime numbers. $\endgroup$ – Mare May 10 '20 at 20:32
  • $\begingroup$ You certainly knew $X_4=\{0,2,4\}$, so is $X_6$ the first case you don't know? $\endgroup$ – YCor May 10 '20 at 20:36
  • $\begingroup$ @YCor Thanks, yes 6 is the first non-trivial case it seems. $\endgroup$ – Mare May 10 '20 at 20:38
  • $\begingroup$ GAP might provide for small $n$ the list of $k$ such that $S_n$ has a transitive solvable subgroup that contains a product of $k$ disjoint transpositions (which is a necessary condition for the existence of such a polynomial with $n-2k$ real roots). Nevertheless for $n=6,8$ this yields no obstruction. $\endgroup$ – YCor May 10 '20 at 20:47

Klueners and Malle have a database of number fields of degree $\leq 19$ that (tries) to include every Galois group and every possible signature. Examining this database shows that for composite $n$ with $4 \leq n \leq 18$, $X_{n} = \{ k : 0 \leq k \leq n \text{ and } k \equiv n \pmod{2} \}$.

EDIT: If $n = 2k$ is even, then $\mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}/k\mathbb{Z}$ contains elements of order $2$ with any desired even number of fixed points. It seems possible that one can obtain the same thing for $n$'s that are multiples of $3$. The smallest $n$ that does not fall into this category is $n = 25$, and using the classification of transitive subgroups of $S_{25}$, one finds that $X_{25} \subseteq \{ 1, 5, 9, 13, 17, 21, 25 \}$.

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    $\begingroup$ So quite surprisingly for primes it tends to be very small and for composite it's as large as possible... until 19. For every such $n$ and $k$ with $n-k$ even you manually checked inside the list the polynomials until finding one with $k$ real roots? or you found a quicker way? $\endgroup$ – YCor May 10 '20 at 21:12
  • $\begingroup$ Thanks, that motivates to experiment a bit also for some n>=20. I will try my luck a bit to see if the pattern continues. $\endgroup$ – Mare May 10 '20 at 22:40

Jensen On the number of real roots of a solvable polynomial includes a proof of:

Loewy’s theorem. Let $K$ be a real number field and $f(X)$ an irreducible polynomial in $K[X]$ of odd degree $n$. If $p$ is the smallest prime divisor of $n$ and the Galois group of $f(X)$ over $K$ is solvable, then $r(f) = 1$ or $n$ or satisfies the inequalities $p ≤ r(f)≤ n−p +1.$


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