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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 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),

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

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,

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

with $a$ as above.

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?

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 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),

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

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?

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 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,

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

Other $p=5m+1$ is 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,

$$D_1 = a^2\,\big( 4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5 - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 \big)^2$$

$$D_2 = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$$

The first is just a $2$nd power, while the latter has a $4$th power. My first thought was to turn the cubic in $b$ into the square $c^2$ (an elliptic curve). What I found to my surprise was it can be solved as the $4$th power $c^4$ (a 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,

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

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

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

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

with $a$ as above.

V. Questions

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

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 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),

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

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,

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

with $a$ as above.

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?

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 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,

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

Other $p=5m+1$ is 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,

$$D_1 = a^2\,\big( 4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5 - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 \big)^2$$

$$D_2 = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$$

The first is just a $2$nd power, while the latter has a $4$th power. 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 be solved as the $4$th power $c^4$,

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

with solution,

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

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

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

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

with $a$ as above.

V. Questions

1. Am I correct in assuming this one-parameter Kondo quintic is now a $5T1$ with order $5$?
2. More importantly, I was trying to make the cubic in $b$ into a square. The mystery for me is, why was it possible to make it a fourth power?

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 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,

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

Other $p=5m+1$ is 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,

$$D_1 = a^2\,\big( 4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5 - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 \big)^2$$

$$D_2 = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$$

The first is just a $2$nd power, while the latter has a $4$th power. My first thought was to turn the cubic in $b$ into the square $c^2$ (an elliptic curve). What I found to my surprise was it can be solved as the $4$th power $c^4$ (a 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,

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

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

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

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

with $a$ as above.

V. Questions

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