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*](http://www.math.uni-duesseldorf.de/~klueners/minimum/)" this seems to be the only one. (Though I was able to find a 2-parameter family of a different form.)



Octic family with Galois group of order 1344?
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). 


