Skip to main content
MathOverflow

Return to Answer

Corrected order for p = 19
Source Link
Tito Piezas III
Tito Piezas III
  • 12.6k
  • 1
  • 39
  • 89

(Not an answer but an addendum that addresses three issues.)

Issue 1

While a detailed answer has explained the method works for the generic cubic, numeric testing suggests any cubic $x^3+\alpha x^2+\beta x+1=0$ will work as long as its discriminant $D$ is a negative square. Is this true?

Issue 2

To find $j,k$ with $j>k$ of the appropriate "surface" $a^jb^k+b^jc^k+c^ja^k$ for a level $p$, what I did was actually solve the system,

$$[a^jb^k,\;b^jc^k,\;c^ja^k] = [u_1,\; u_2,\; u_3]$$

for the "seed" cubic roots $a,b,c$ (easily done in Mathematica). For example, for $p=37$ and using $j=7$, only for $k = 3$, we get,

$$a = \left(\frac{u_1^{49}\,u_3^9}{u_2^{21}}\right)^{1/370}$$

So I figure if the fractional power contains a multiple of $p=37$, then this is the correct $j,k$. It seems then,

$$a^7b^3+b^7c^3+c^7a^3\quad\text{(deg 37)}$$ $$\; a^7b+b^7c+c^7a\quad\quad \text{(deg 43)}$$ $$a^9b^4+b^9c^4+c^9a^4\;\quad \text{(deg 61)}$$ $$a^9b^2+b^9c^2+c^9a^2\quad \text{(deg 67)}$$ $$a^9b+b^9c+c^9a\quad\quad \text{(deg 73)}$$

Using $j=8$ with $k<j$ just yields multiples of lower primes.

Issue 3

For p = 7, constructing the two septics was easy. The Mathematica command Table[Mod[k^2, 7], {k, 1, 3}] gives ($1,4,2$) so with the roots $u_i$ of the first cubic,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_7^{k}\,u_1^{1/7}+\zeta_7^{4k}\,u_2^{1/7}+\zeta_7^{2k}\,u_3^{1/7}\big)\Big)$$

yields the first septic, while the roots $v_i$ of the second cubic is for the partner septic.

For p = 13, it gives ($1, \color{blue}4, 9, 3, \color{blue}{12}, \color{blue}{10}$) so one has to find the correct triplets. By trial-and-error,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_{13}^{k}\,u_1^{1/13}+\zeta_{13}^{9k}\,u_2^{1/13}+\zeta_{13}^{3k}\,u_3^{1/13}\big)\Big)$$$$P(x) = \prod_{k=0}^{12} \Big(x-\big(\zeta_{13}^{k}\,u_1^{1/13}+\zeta_{13}^{9k}\,u_2^{1/13}+\zeta_{13}^{3k}\,u_3^{1/13}\big)\Big)$$ $$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_{13}^{\color{blue}{4k}}\,u_1^{1/13}+\zeta_{13}^{\color{blue}{10k}}\,u_2^{1/13}+\zeta_{13}^{\color{blue}{12k}}\,u_3^{1/13}\big)\Big)$$$$P(x) = \prod_{k=0}^{12} \Big(x-\big(\zeta_{13}^{\color{blue}{4k}}\,u_1^{1/13}+\zeta_{13}^{\color{blue}{10k}}\,u_2^{1/13}+\zeta_{13}^{\color{blue}{12k}}\,u_3^{1/13}\big)\Big)$$

so now there are two ways to generate the same 13-deg. (Using $v_i$ gives the partner 13-deg.)

For p = 19, it gives ($1, \color{blue}4, \color{red}5, \color{blue}6, 7, \color{blue}9, 11, \color{red}{16}, \color{red}{17}$$1, \color{blue}4, \color{blue}9, \color{red}{16}, \color{blue}6, \color{red}{17}, 11, 7, \color{red}5$) so there are three ways to generate the same 19-deg. And so on. Is there a way to find the correct triplets without trial-and-error?

(Not an answer but an addendum that addresses three issues.)

Issue 1

While a detailed answer has explained the method works for the generic cubic, numeric testing suggests any cubic $x^3+\alpha x^2+\beta x+1=0$ will work as long as its discriminant $D$ is a negative square. Is this true?

Issue 2

To find $j,k$ with $j>k$ of the appropriate "surface" $a^jb^k+b^jc^k+c^ja^k$ for a level $p$, what I did was actually solve the system,

$$[a^jb^k,\;b^jc^k,\;c^ja^k] = [u_1,\; u_2,\; u_3]$$

for the "seed" cubic roots $a,b,c$ (easily done in Mathematica). For example, for $p=37$ and using $j=7$, only for $k = 3$, we get,

$$a = \left(\frac{u_1^{49}\,u_3^9}{u_2^{21}}\right)^{1/370}$$

So I figure if the fractional power contains a multiple of $p=37$, then this is the correct $j,k$. It seems then,

$$a^7b^3+b^7c^3+c^7a^3\quad\text{(deg 37)}$$ $$\; a^7b+b^7c+c^7a\quad\quad \text{(deg 43)}$$ $$a^9b^4+b^9c^4+c^9a^4\;\quad \text{(deg 61)}$$ $$a^9b^2+b^9c^2+c^9a^2\quad \text{(deg 67)}$$ $$a^9b+b^9c+c^9a\quad\quad \text{(deg 73)}$$

Using $j=8$ with $k<j$ just yields multiples of lower primes.

Issue 3

For p = 7, constructing the two septics was easy. The Mathematica command Table[Mod[k^2, 7], {k, 1, 3}] gives ($1,4,2$) so with the roots $u_i$ of the first cubic,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_7^{k}\,u_1^{1/7}+\zeta_7^{4k}\,u_2^{1/7}+\zeta_7^{2k}\,u_3^{1/7}\big)\Big)$$

yields the first septic, while the roots $v_i$ of the second cubic is for the partner septic.

For p = 13, it gives ($1, \color{blue}4, 9, 3, \color{blue}{12}, \color{blue}{10}$) so one has to find the correct triplets. By trial-and-error,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_{13}^{k}\,u_1^{1/13}+\zeta_{13}^{9k}\,u_2^{1/13}+\zeta_{13}^{3k}\,u_3^{1/13}\big)\Big)$$ $$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_{13}^{\color{blue}{4k}}\,u_1^{1/13}+\zeta_{13}^{\color{blue}{10k}}\,u_2^{1/13}+\zeta_{13}^{\color{blue}{12k}}\,u_3^{1/13}\big)\Big)$$

so now there are two ways to generate the same 13-deg. (Using $v_i$ gives the partner 13-deg.)

For p = 19, it gives ($1, \color{blue}4, \color{red}5, \color{blue}6, 7, \color{blue}9, 11, \color{red}{16}, \color{red}{17}$) so there are three ways to generate the same 19-deg. And so on. Is there a way to find the correct triplets without trial-and-error?

(Not an answer but an addendum that addresses three issues.)

Issue 1

While a detailed answer has explained the method works for the generic cubic, numeric testing suggests any cubic $x^3+\alpha x^2+\beta x+1=0$ will work as long as its discriminant $D$ is a negative square. Is this true?

Issue 2

To find $j,k$ with $j>k$ of the appropriate "surface" $a^jb^k+b^jc^k+c^ja^k$ for a level $p$, what I did was actually solve the system,

$$[a^jb^k,\;b^jc^k,\;c^ja^k] = [u_1,\; u_2,\; u_3]$$

for the "seed" cubic roots $a,b,c$ (easily done in Mathematica). For example, for $p=37$ and using $j=7$, only for $k = 3$, we get,

$$a = \left(\frac{u_1^{49}\,u_3^9}{u_2^{21}}\right)^{1/370}$$

So I figure if the fractional power contains a multiple of $p=37$, then this is the correct $j,k$. It seems then,

$$a^7b^3+b^7c^3+c^7a^3\quad\text{(deg 37)}$$ $$\; a^7b+b^7c+c^7a\quad\quad \text{(deg 43)}$$ $$a^9b^4+b^9c^4+c^9a^4\;\quad \text{(deg 61)}$$ $$a^9b^2+b^9c^2+c^9a^2\quad \text{(deg 67)}$$ $$a^9b+b^9c+c^9a\quad\quad \text{(deg 73)}$$

Using $j=8$ with $k<j$ just yields multiples of lower primes.

Issue 3

For p = 7, constructing the two septics was easy. The Mathematica command Table[Mod[k^2, 7], {k, 1, 3}] gives ($1,4,2$) so with the roots $u_i$ of the first cubic,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_7^{k}\,u_1^{1/7}+\zeta_7^{4k}\,u_2^{1/7}+\zeta_7^{2k}\,u_3^{1/7}\big)\Big)$$

yields the first septic, while the roots $v_i$ of the second cubic is for the partner septic.

For p = 13, it gives ($1, \color{blue}4, 9, 3, \color{blue}{12}, \color{blue}{10}$) so one has to find the correct triplets. By trial-and-error,

$$P(x) = \prod_{k=0}^{12} \Big(x-\big(\zeta_{13}^{k}\,u_1^{1/13}+\zeta_{13}^{9k}\,u_2^{1/13}+\zeta_{13}^{3k}\,u_3^{1/13}\big)\Big)$$ $$P(x) = \prod_{k=0}^{12} \Big(x-\big(\zeta_{13}^{\color{blue}{4k}}\,u_1^{1/13}+\zeta_{13}^{\color{blue}{10k}}\,u_2^{1/13}+\zeta_{13}^{\color{blue}{12k}}\,u_3^{1/13}\big)\Big)$$

so now there are two ways to generate the same 13-deg. (Using $v_i$ gives the partner 13-deg.)

For p = 19, it gives ($1, \color{blue}4, \color{blue}9, \color{red}{16}, \color{blue}6, \color{red}{17}, 11, 7, \color{red}5$) so there are three ways to generate the same 19-deg. And so on. Is there a way to find the correct triplets without trial-and-error?

Source Link
Tito Piezas III
Tito Piezas III
  • 12.6k
  • 1
  • 39
  • 89

(Not an answer but an addendum that addresses three issues.)

Issue 1

While a detailed answer has explained the method works for the generic cubic, numeric testing suggests any cubic $x^3+\alpha x^2+\beta x+1=0$ will work as long as its discriminant $D$ is a negative square. Is this true?

Issue 2

To find $j,k$ with $j>k$ of the appropriate "surface" $a^jb^k+b^jc^k+c^ja^k$ for a level $p$, what I did was actually solve the system,

$$[a^jb^k,\;b^jc^k,\;c^ja^k] = [u_1,\; u_2,\; u_3]$$

for the "seed" cubic roots $a,b,c$ (easily done in Mathematica). For example, for $p=37$ and using $j=7$, only for $k = 3$, we get,

$$a = \left(\frac{u_1^{49}\,u_3^9}{u_2^{21}}\right)^{1/370}$$

So I figure if the fractional power contains a multiple of $p=37$, then this is the correct $j,k$. It seems then,

$$a^7b^3+b^7c^3+c^7a^3\quad\text{(deg 37)}$$ $$\; a^7b+b^7c+c^7a\quad\quad \text{(deg 43)}$$ $$a^9b^4+b^9c^4+c^9a^4\;\quad \text{(deg 61)}$$ $$a^9b^2+b^9c^2+c^9a^2\quad \text{(deg 67)}$$ $$a^9b+b^9c+c^9a\quad\quad \text{(deg 73)}$$

Using $j=8$ with $k<j$ just yields multiples of lower primes.

Issue 3

For p = 7, constructing the two septics was easy. The Mathematica command Table[Mod[k^2, 7], {k, 1, 3}] gives ($1,4,2$) so with the roots $u_i$ of the first cubic,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_7^{k}\,u_1^{1/7}+\zeta_7^{4k}\,u_2^{1/7}+\zeta_7^{2k}\,u_3^{1/7}\big)\Big)$$

yields the first septic, while the roots $v_i$ of the second cubic is for the partner septic.

For p = 13, it gives ($1, \color{blue}4, 9, 3, \color{blue}{12}, \color{blue}{10}$) so one has to find the correct triplets. By trial-and-error,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_{13}^{k}\,u_1^{1/13}+\zeta_{13}^{9k}\,u_2^{1/13}+\zeta_{13}^{3k}\,u_3^{1/13}\big)\Big)$$ $$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_{13}^{\color{blue}{4k}}\,u_1^{1/13}+\zeta_{13}^{\color{blue}{10k}}\,u_2^{1/13}+\zeta_{13}^{\color{blue}{12k}}\,u_3^{1/13}\big)\Big)$$

so now there are two ways to generate the same 13-deg. (Using $v_i$ gives the partner 13-deg.)

For p = 19, it gives ($1, \color{blue}4, \color{red}5, \color{blue}6, 7, \color{blue}9, 11, \color{red}{16}, \color{red}{17}$) so there are three ways to generate the same 19-deg. And so on. Is there a way to find the correct triplets without trial-and-error?