The irreducible but solvable octic,


$$x^8-x^7+29x^2+29=0\tag{1}\label{1}$$


was first mentioned by Igor Schein in this 1999 [sci.math post](http://www.math.niu.edu/%7Erusin/known-math/99/galois_8) ([Wayback Machine](https://web.archive.org/web/20110530123122/http://www.math.niu.edu/%7Erusin/known-math/99/galois_8)). This does not factor over a quadratic or quartic extension, but over a 7th deg one. It can also be nicely solved using the $\color{blue}{29}$th root of unity. Let $\omega = \exp(2\pi i /29)$ then define,


\begin{gather}
y\_k = \omega^{k}+\omega^{12k}+\omega^{17k}+\omega^{28k}\tag{2}\label{2} \\
z\_k = 4(y\_k^3+y\_k^2-9y\_k-4)(y\_k^2-2)(y\_k-1)+9\tag{3}\label{3}
\end{gather}


then I found a pair of octic roots as,


\begin{gather\*}
x = \frac{1\color{red}{-}\sqrt{z\_{1}}+\sqrt{z\_{2}}+\sqrt{z\_{4}}+\sqrt{z\_{8}}+\sqrt{z\_{16}}+\sqrt{z\_{32}}+\sqrt{z\_{64}}}{8} \approx 1.79106+0.8286\,i\dots \\
x = \frac{1+\sqrt{z\_{1}}\color{red}{-}\sqrt{z\_{2}}\color{red}{-}\sqrt{z\_{4}}+\sqrt{z\_{8}}+\sqrt{z\_{16}}\color{red}{-}\sqrt{z\_{32}}+\sqrt{z\_{64}}}{8} \approx 1.79106-0.8286\,i\dots
\end{gather\*}


and the other pairs using appropriate signs of the square roots. Note that the seven $y\_k$ and $z\_k$ are roots of two different 7th-deg eqns with integer coefficients, with the $z\_k$ being,


$$\small{z^7 - 7z^6 - 2763z^5 - 19523 z^4 + 1946979z^3 + 34928043 z^2 + 119557031z - 3247^2=0}$$


and \eqref{3} is the 6th-deg Tschirnhausen transformation between the $y\_k$ and $z\_k$. (In an earlier edit, I used an alternative expression for $z\_k$ by P. Montgomery found in the sci.math link, but I like this one better.)


***Question***: Does anyone know why \eqref{1} has such a *simple* form, and if we can find other irreducible but solvable octics with the same Galois group involving a $p$th root of unity for other prime $p$? (For some reason, this does not appear in the Kluener's [database of number fields for 8T25](http://reh.math.uni-duesseldorf.de/cgi-klueners/groups3.pl?deg=8&t=25&zer=&bet=&teil=&noteil=&maxteil=&numprime=).)



On the solvable octic $x^8-x^7+29x^2+29=0$
To answer your second question, there are soluble octics with the same Galois group involving
other $p$th roots of unity. Take $p\equiv 1\mod 7$, and $K={\mathbb Q}(\alpha)$
the unique degree 7 extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta\_p)$. E.g. take
$\alpha=\sum\_i \zeta\_p^i$ where $i$ ranges over all seventh powers in ${\mathbb F\_p}$.


If $f(x)$ is the minimal polynomial of $\alpha$ (degree 7), then $f(x^2)$ is the minimal
polynomial of $\sqrt\alpha$, which defines, generally, a 'random' quadratic extension of $K$.
That is, its Galois group over ${\mathbb Q}$ is
$$
 G=C\_2\wr C\_7\cong C\_2^7:C\_7.
$$
Viewing $C\_2^7$ as a 7-dimensional representation
of $C\_7$ over ${\mathbb F}\_2$, it decomposes as a 1-dimensional (trivial) representation plus
two distinct 3-dimensional ones. (The reason for this is that $2^3\equiv 1\mod 7$.) Factor out
$C\_2^4\triangleleft G$, which is one of those
plus the trivial one. This gives a
Galois group $C\_2^3:C\_7$ that you are after,
and a subgroup $C\_7$ in it cuts out the required octic field.


Here is a Magma code that can be used in the
[Magma online calculator](http://magma.maths.usyd.edu.au/calc/)
to get such an octic:



```
p:=43;        // or some other p = 1 mod 7

K<z>:=CyclotomicField(p);
alpha:=&+[z^i: i in [1..p] | IsPower(GF(p)!i,7)];    

R<x>:=PolynomialRing(Rationals());
f:=Evaluate(MinimalPolynomial(alpha),x^2);
K:=NumberField(f);
assert exists(a){a: a in ArtinRepresentations(K) | #Kernel(Character(a)) eq 16};
F:=Field(Minimize(a));
DefiningPolynomial(F);

```

You can also stick in a Tschirnhaus transformation, say,



```
alpha:=alpha^3+alpha+1;

```

in the 5th line to vary the generator of $K$ — in this way you get all possible
$C\_2^3:C\_7$-extensions involving $p$th roots of unity.


For your questions in the [comments](https://mathoverflow.net/questions/145145/on-the-solvable-octic-x8-x729x229-0#comment374605_145145), the roots might be real or complex, and the constant term
may or may not be a square — this depends on whether $\alpha$ is chosen to be totally real,
and on the way 'Minimize' works; you can always use an additional Tschirnhaus transformation
to modify the final output or Pari's `polredabs` function to try and get the coefficients smaller.


I do not know the reason why for $p=29$ there is such an elegant octic, this is very curious. It is a bit like the Trinks polynomial $x^7-7x+3$ with Galois group $\operatorname{PSL}(2,7)$, and I wonder whether simple polynomials having interesting Galois group is such a statistical blip, or there is a reason behind it.


