I've always wondered if the DeMoivre method to generate an algebraic number $x_p$,

$$x_p = u_1^{1/p}+u_2^{1/p}$$

of degree $p$ using only quadratic roots $u_i$ could be generalized using cubic roots $v_i$,

$$x_p = v_1^{1/p}+v_2^{1/p}+v_3^{1/p}$$

I serendipitously found a method which works for prime $p=6m+1$, the clue being the Klein quartic for $p=7$. The surfaces I used starts with that,

$$a^3b+b^3c+c^3a\quad \text{(deg 7)}$$ $$\;a^4b+b^4c+c^4a\;\quad \text{(deg 13)}$$ $$a^5b^2+b^5c^2+c^5a^2\quad \text{(deg 19)}$$

and so on. First, given the generic cubic for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

with roots $a,b,c.$

I. Level 7

Using the roots $a,b,c$ of the generic cubic, then,

$$x_7 = (a^3b)^{1/7}+(b^3c)^{1/7}+(c^3a)^{1/7}$$ $$x_7^{'} = (a^3c)^{1/7}+(b^3a)^{1/7}+(c^3b)^{1/7}$$

properly chosen, are the roots of,

\begin{align} x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= 28x - (n^2 + 5n + 9)\\ x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= -7(n^2 - 3n + 5)x + (n^3 - 4n^2 + 4n - 9) \end{align}

The method generates pairs of solvable equations that curiously differ only by 2 coefficients for $p=7$, by 4 coefficients for $p=13$, etc.

II. Level 13

Using the same $a,b,c$, then,

$$x_{13} = (a^4b)^{1/13}+(b^4c)^{1/13}+(c^4a)^{1/13}$$ $$x_{13}^{'} = (a^4c)^{1/13}+(b^4a)^{1/13}+(c^4b)^{1/13}$$

are the roots of,

$$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = A$$ $$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = B$$

where,

$$A = -442x^4 + 13 (9 + 15 n + n^2) x^2 - 13 (36 - 3 n + n^2) x + (189 - 72 n + 15 n^2 - n^3)$$ $$B = 26 (1 - 6 n + 2 n^2) x^4 + 13 (9 + 6 n + 4 n^2 - n^3) x^2 + 13 (-3 + 2 n)^2 x + (18 + n^2) (12 - 5 n + n^2)$$

hence differs only by 4 coefficients.

III. Level 19

Still using the same $a,b,c$, then,

$$x_{19} = (a^5b^2)^{1/19}+(b^5c^2)^{1/19}+(c^5a^2)^{1/19}$$ $$x_{19}^{'} = (a^5c^2)^{1/19}+(b^5a^2)^{1/19}+(c^5b^2)^{1/19}$$

are the roots of two $19$-deg equations that differ only by several coefficients. (A parametric example can be given, but will clutter up the post.)

IV. Levels 31 and 37

$$x_{31} = (a^6b)^{1/31}+(b^6c)^{1/31}+(c^6a)^{1/31}$$

$$x_{37} = (a^7b^3)^{1/37}+(b^7c^3)^{1/37}+(c^7a^3)^{1/37}$$

and so on.

V. Questions

1. Why, starting with the Klein quartic and the roots of the generic cubic, does the method work?
2. By trial and error, I've found the surfaces for $p=43,61,67,73$. If the method works, can we predict these in advance from first principles?

I've always wondered if the DeMoivre method to generate an algebraic number $x_p$,

$$x_p = u_1^{1/p}+u_2^{1/p}$$

of degree $p$ using only quadratic roots $u_i$ could be generalized using cubic roots $v_i$,

$$x_p = v_1^{1/p}+v_2^{1/p}+v_3^{1/p}$$

I serendipitously found a method which works for prime $p=6m+1$, the clue being the Klein quartic for $p=7$. The surfaces I used starts with that,

$$a^3b+b^3c+c^3a\quad \text{(deg 7)}$$ $$\;a^4b+b^4c+c^4a\;\quad \text{(deg 13)}$$ $$a^5b^2+b^5c^2+c^5a^2\quad \text{(deg 19)}$$

and so on. First, given the generic cubic for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

with discriminant $D=-(n^2-3n+9)^2,$ hence all roots $a,b,c$ are real.

I. Level 7

Using the roots $a,b,c$ of the generic cubic, then,

$$x_7 = (a^3b)^{1/7}+(b^3c)^{1/7}+(c^3a)^{1/7}$$ $$x_7^{'} = (a^3c)^{1/7}+(b^3a)^{1/7}+(c^3b)^{1/7}$$

properly chosen, are the roots of,

\begin{align} x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= 28x - (n^2 + 5n + 9)\\ x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= -7(n^2 - 3n + 5)x + (n^3 - 4n^2 + 4n - 9) \end{align}

The method generates pairs of solvable equations that curiously differ only by 2 coefficients for $p=7$, by 4 coefficients for $p=13$, etc.

II. Level 13

Using the same $a,b,c$, then,

$$x_{13} = (a^4b)^{1/13}+(b^4c)^{1/13}+(c^4a)^{1/13}$$ $$x_{13}^{'} = (a^4c)^{1/13}+(b^4a)^{1/13}+(c^4b)^{1/13}$$

are the roots of,

$$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = A$$ $$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = B$$

where,

$$A = -442x^4 + 13 (9 + 15 n + n^2) x^2 - 13 (36 - 3 n + n^2) x + (189 - 72 n + 15 n^2 - n^3)$$ $$B = 26 (1 - 6 n + 2 n^2) x^4 + 13 (9 + 6 n + 4 n^2 - n^3) x^2 + 13 (-3 + 2 n)^2 x + (18 + n^2) (12 - 5 n + n^2)$$

hence differs only by 4 coefficients.

III. Level 19

Still using the same $a,b,c$, then,

$$x_{19} = (a^5b^2)^{1/19}+(b^5c^2)^{1/19}+(c^5a^2)^{1/19}$$ $$x_{19}^{'} = (a^5c^2)^{1/19}+(b^5a^2)^{1/19}+(c^5b^2)^{1/19}$$

are the roots of two $19$-deg equations that differ only by several coefficients. (A parametric example can be given, but will clutter up the post.)

IV. Levels 31 and 37

$$x_{31} = (a^6b)^{1/31}+(b^6c)^{1/31}+(c^6a)^{1/31}$$

$$x_{37} = (a^7b^3)^{1/37}+(b^7c^3)^{1/37}+(c^7a^3)^{1/37}$$

and so on.

V. Questions

1. Why, starting with the Klein quartic and the roots of the generic cubic, does the method work?
2. By trial and error, I've found the surfaces for $p=43,61,67,73$. If the method works, can we predict these in advance from first principles?