**I. Kondo-Brumer quintic**


The deceptively simple solvable quintic,


$$x^5 + (a - 3)x^4 + (-a + b + 3)x^3 + (a^2 - a - 1 - 2b)x^2 + b x + a=0$$


is quite important for *imaginary quadratic fields*. For example, let $a=1, b=0,$ and it becomes,


$$x^5-2x^4+2x^3-x^2+1=0$$


which is a [Weber class polynomial](https://en.wikipedia.org/wiki/Weber_modular_function) and a solution is,


$$x= -e^{-2\pi i/48}\frac{\sqrt{2}\,\eta(2\tau)}{\quad\eta(\tau)}\approx -0.5764$$


with Dedekind eta function $\eta(\tau)$ and $\tau=\frac{1+\sqrt{-47}}2.$ Other $\sqrt{-d}$ with class number $h(d) = 5m$ are also possible.


**II. Lehmer quintic**


This is given by,


$$y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y +1=0$$


and, on the other hand, is quite important for *real fields*. For example. let $n=-1$ then,


$$y^5+y^4-4y^3-3y^2+3y+1=0$$


and a solution is,


$$y = 2\cos(2\pi/11) \approx 1.6825$$


Of course, other $p=5m+1$ are also possible.


**III. Subset of a, b**


The two quintics seem to be radically different, the first generally a $5T2$ with order $10$, and the second a $5T1$ with order $5$. But it turns out (using a subset of $a,b$) we can transform the former into the latter. Their discriminants are,


$$D\_1 = a^2\,\big( 4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5 - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 \big)^2$$


$$D\_2 = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$$


My first thought was to turn the cubic in $b$ into the square $c^2$ (so an elliptic curve). What I found to my surprise was it can also be solved as the $4$th power $c^4$ (a [*superelliptic curve*](https://en.wikipedia.org/wiki/Superelliptic_curve)),


$$\small{(4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5) - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 = \color{blue}{c^4}}$$


with solution,


$$\begin{align}
a &= -(7 + 10 n + 5 n^2 + n^3)\\
b &= -20 - 5 n + 10 n^2 + 12 n^3 + 5 n^4 + n^5\\
c &=\, 25 + 25 n + 15 n^2 + 5 n^3 + n^4
\end{align}$$


Incidentally, the $a,b,c$ have the nice linear relationship,


$$3a-b+nc = -1$$


So the new discriminant of the one-parameter Kondo quintic for these special $a,b$ neatly becomes,


$$D\_1(\text{new}) = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^{\color{blue}8}$$


and has an $8$th power.


**IV. Tschirnhausen**


To complete the transformation, there is a quartic Tschirnhausen between the one-parameter Kondo quintic (in $x$) and the Lehmer (in $y$) given by,


$$ax = y^4 - (4 + 3 n) y^3 - (8 + 13 n + 13 n^2 + 6 n^3 + n^4) y^2 + (30 + 60 n + 54 n^2 + 28 n^3 + 8 n^4 + n^5) y + (2 + n)$$


with $a$ as above. Incidentally, if we are to express the quartic Tschirnhausen in the form,


$$ax = y^4+py^3+qy^2+ry+s$$


then the coefficients $a,p,q,r,s$ also have a nice relationship,


$$3a - p + q s + r + s = - 1$$


**V. Questions**


1. Am I correct in assuming this one-parameter Kondo quintic is now a $5T1$ with order $5$?
2. More importantly, since I was trying to make the cubic in $b$ into a square, a mystery to me is *why was it possible and easier to make it a 4th power and solve it as a superelliptic curve?*



Transforming the Kondo quintic $5T2$ into the Lehmer quintic $5T1$?
Regarding your first question, the answer is yes. In fact, your quartic Tschirnhausen formula shows that. In particular, this shows that the one-parameter Kondo quintic has a root in the splitting field of the Lehmer quintic, which is a degree $5$ Galois extension of $\mathbb{Q}(n)$. Because it's Galois, it's normal, and since it's normal, the one-parameter Kondo quintic must split in this field.
This implies that the Galois group of the one-parameter Kondo quintic is $5T1$. (If you want, you can also compute the Galois group of a polynomial over a one-variable function field with Magma and independently check this.)


Your second question is a bit trickier - the why questions in mathematics can be hard to answer. If you look at the elliptic curve that comes from setting the discriminant to be a square, namely
$$ y^{2} = x^{3} + (1-30a+a^{2})x^{2} - 8(7a + 17a^{2} - 12a^{3})x + 16(4a - 91a^{2} + 40a^{3} + 4a^{4} - 4a^{5}),
$$
then this elliptic curve has $(n^{4} + 5n^{3} + 15n^{2} + 25n + 25)^{5}$ as a factor of its discriminant, which is an unusually high power. Moreover, at this place of bad reduction (which corresponds to choosing $n$ in $\mathbb{Q}(\zeta\_{5})$), the reduction is split multiplicative of type $I\_{5}$. In particular, the component group $E/E\_{0}$ is cyclic of order $5$. The solution $a$, $b$, $c$ you specified gives an integral point $P$ on this elliptic curve (that is, the coordinates of $P$ are polynomials in $n$, rather than rational functions). It is somewhat remarkable in that $P$, $2P$ and $3P$ are all integral points as well. One consequence of the reduction properties at this place is that $n^{4} + 5n^{4} + 15n^{3} + 25n + 25$ divides the $y$-coordinate of $mP$ to the power $2$ whenever $m \equiv 1, 4 \pmod{5}$. This general phenomena among other things forces $(n^{4} + 5n^{4} + 15n^{3} + 25n + 25)^{2}$ to divide the $y$-coordinate of $P$. These things are sensitive to the particular set up. I'm not sure if you first tried to find a parametric family of $5T1$ specializations of the Brumer-Kondo quintic by setting $a = 7 + 10n + 5n^{2} + n^{3}$ (the negative of what you chose). You may have quickly figured out that this doesn't work: in this situation the Mordell-Weil rank of the elliptic curve over $\mathbb{Q}(n)$ is zero, so you do not get a parametric family of $5T1$ specializations.


