Question

Does the octic,

$\tag{1} x^8+3x^7-15x^6-29x^5+79x^4+61x^3+29x+16 = nx^2$

for any constant n have Galois group of order 1344? Its discriminant D is a perfect square,

$D = (1728n^4-341901n^3-11560361n^2+3383044089n+28121497213)^2$

Surely (1) is not an isolated result. How easy is it to find another family with the same Galois group and same form,

$\tag{2} \text{octic poly in}\ x = nx^2$

Perusing Kluener's "A Database For Number Fields" this seems to be the only one. (Though I was able to find a 2-parameter family of a different form.)

Answer 1

Maple gives Galois group $H$ of your polynomial $f=f_n$ as the subgroup of $S_8$ generated by these permutations: $$(1 2)(5 6), (1 2 3)(4 6 5), (1 2 6 3 4 5 7), (1 8)(2 3)(4 5)(6 7), (2 8)(1 3)(4 6)(5 7), (4 8)(1 5)(2 6)(3 7).$$ That can be easily proved by using the standard technique of Galois theory assuming that $n$ is such that your polynomial is irreducible (I think that the only exceptional values are 17 and 145 as in Alastair's answer). See Van der Waerden's book, for example: consider the polynomial in $9$ variables $$g(x,x_1,...,x_8)=\prod_{\sigma\in S_8} (x-x_{\sigma 1}a_1-...-x_{\sigma 8}a_8) $$ where $a_1,...,a_8$ are the roots of your polynomial $f=f_n$, $\sigma$ runs over all permutations of $S_8$. The coefficients of $g$ are symmetric polynomials in $a_i$, hence polynomials in the coefficients of $f_n$. Now show that the products of factors corresponding to permutations from $H$ form an irreducible factor of $g$. Hence $H$ is the Galois group of $f_n$ (again, assuming $f_n$ is irreducible).

Answer 2

More of a comment than an answer, but there are exceptions for $n = 17$ and $n = 145$, where the Galois group is simple of order 168. This is all the exceptions for $|n| < 2\times10^5$.

Answer 3

In arxiv.org/abs/1209.5300 I give the following polynomial with the same Galois group over Q(t): $(y+1)(y^7-y^6-11y^5+y^4+41y^3+25y^2-34y-29)-t(2y+3)^2$. This has $(6912t^4-3456t^3-95472t^2+23976t-1417)^2$ for a discriminant. This doesn't quite answer the second question, but substituting $y \mapsto (x-3)/2$, $t \mapsto s/256$ gives $(x-1) (x^7-23 x^6+181 x^5-547 x^4+515 x^3-325 x^2+71 x-1)-s x^2$, with discriminant $256 (27 s^4-3456 s^3-24440832 s^2+1571291136 s-23773315072)^2$, which does.