4

We have, $$x^3+x^2-2x-1=0,\quad\quad x =\sum_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, (6^k)}{7}\Bigr)\\ x^3+x^2-4x+1=0,\quad\quad x =\sum_{k=1}^{4}\,\exp\Bigl(\tfrac{2\pi\, i\, (5^k)}{13}\Bigr)\\ x^3+x^2-6x-7=0,\quad\quad x =\sum_{k=1}^{6}\,\exp\Bigl(\tfrac{2\pi\, i\, (8^k)}{19}\Bigr)$$ and so on. It seems the general formula is, given a prime $p=6m+1$ and the unique solution $u,v$ to, $$u^2+27v^2=4p$$ then,

$$x^3+x^2-2mx+\frac{1-3p\pm p u}{27}=0\tag1$$

for the appropriate sign $\pm$ and where $x =\sum_{k=1}^{2m}\,\exp\Bigl(\tfrac{2\pi\, i\, (\color{blue}c^k)}{p}\Bigr)$ and appropriate integer $\color{blue}c$. For example, for $p=7$, we have $$(-1)^2+27\times1^2=4\times7$$ and using $u=-1$ on formula $(1)$, we recover the first example.

Q: Is it possible to find a formula for the sign $\pm$ and/or $\color{blue}c\,$ as a function of $p$?

$\color{blue}{Update}$: It seems the answer to the part about the sign is yes. While checking the OEIS for the sequence of solutions $u=1, 5, 7, 4, 11, 8,\dots$, it turns out there is signed version using the Kronecker symbol(?). Thus,

$$x^3+x^2-2mx+\frac{1-3p- p\,u'}{27}=0\tag2$$

where $u'=1, -5, 7, 4, -11, -8,\dots$ and given by A123489.

  • Choose the sign according to $u$ mod $3$. The discriminants of your cubics are $v^2p^2$, so I expect this to be false whenever $v^2 \neq 1$. Have you checked the roots for $p=31$ ? – mercio Nov 20 '16 at 11:20
  • @mercio: I've checked the formula up to $p=163$ and the formula works with the sequence of signs as $$-1, 1, -1, -1, 1, 1, -1, 1, -1, 1, -1, -1, 1, 1, 1, -1, 1, -1,\dots$$ – Tito Piezas III Nov 20 '16 at 11:36
  • @mercio: I've figured out how to choose the sign. It is given by a sum using the Kronecker symbol (?). Kindly see update above. – Tito Piezas III Nov 20 '16 at 12:11
  • Well as I was saying the Kronecker symbol picks for you the unique solution with $u = 1 \pmod 3$ and not the unique solution with $u > 0$. (and obviously you need $u' = 1$ mod $3$ or else you can't divide the thing by $27$) – mercio Nov 20 '16 at 12:13
  • @mercio: Would you know how to find $c$? This MO post about the quintic version might help wherein a formula for $c$ was given in Rouse's answer – Tito Piezas III Nov 20 '16 at 12:18
  • Well $c$ can be any element of $\Bbb F_p^*$ of order $(p-1)/3$. – mercio Nov 20 '16 at 12:21
  • 1
    You're asking about what are called "period polynomials". Gauss worked this stuff out in the Disquisitiones. A more recent treatment is in Storer's book, Cyclotomy and Difference Sets. – Gerry Myerson Nov 20 '16 at 12:22
  • @GerryMyerson:The quintic version has $x^5+x^4-4mx^3+ax^2+bx+c=0$ as pointed out by Jagy in this post. If for cubics the Diophantine equation $F(u,v)$ involved is $u^2+27v^2=4p$, would you know what it is for quintics? And can we express $a,b,c$ in terms of $p$ and $F(u,v)$? – Tito Piezas III Nov 20 '16 at 12:28
  • as for my first comment, I confused the discriminant of the polynomial (related to the volume of $\langle 1, x, x^2 \rangle$) with the discriminant of the extension (the volume of $\langle 1, x_1, x_2 \rangle$). A computation should show that $vx_2 \in \Bbb Z[x_1]$ and solve the discrepancy. – mercio Nov 20 '16 at 12:37
  • @mercio: Can you collate your comments perhaps with more detail as an answer so I can accept it? – Tito Piezas III Nov 20 '16 at 13:20
  • There is a diophantine equation involved in the quintic. It might be in Storer's book, I'm not sure, but I'm sure it's more complicated than what comes up in the cubic and quartic. What happened when you typed "period polynomial" and "quintic" into Google? – Gerry Myerson Nov 20 '16 at 21:37
  • @GerryMyerson there is a short presentation of the Gauss material in Galois theory by David A. Cox. That's how I made the lists. Probably beyond me, but everything is determined once the prime and a primitive root are chosen; the primitive root does not matter in the end, but enters into calculating everything. Anyway, requested the Storer book you mention. I guess I don't know what you mean by "There is a diophantine equation involved in the quintic" but don't worry about it, I enjoyed getting the programs to work. – Will Jagy Nov 20 '16 at 23:31
  • 1
    Akinari Hoshi, Explicit lifts of quintic Jacobi sums and period polynomials for $F_q$, Proc. Japan Acad. Ser. A Math. Sci. 82 (2006), no. 7, 87–92, MR2265605 may have some material of interest on the quintic. – Gerry Myerson Nov 21 '16 at 00:01
  • @GerryMyerson this one cites you https://arxiv.org/pdf/1501.07507.pdf – Will Jagy Nov 21 '16 at 00:28
  • 2
    @Will, fancy that. The paper they cite came from a chapter of my PhD thesis. – Gerry Myerson Nov 21 '16 at 02:00
  • @GerryMyerson: Sorry for the late reply. Hoshi's paper or any of Google's results didn't mention any Diophantine equation for the quintic case, but I found my own. Kindly see this MO post. – Tito Piezas III Nov 23 '16 at 14:05

2 Answers2

2

If $p = 3n+1$ and $u,v$ satisfy $u^2+27v^2 = 4p$ with $u \equiv 1 \pmod 3$

then taking this relation modulo $27$ we get $p \equiv 7u^2$, and then $1-3p-pu \equiv 1-21u^2-7u^3 \equiv 7(1-u)^3 - 12(1-u)^2 + 9(1-u) \equiv 0$ because $1-u \equiv 0 \pmod 3$.

Then if we call $w = \frac 1 {27} (1-3p-pu)$, $w$ is an integer.
If $p$ is odd then $w = u = v \pmod 2$ and $n$ is even.

The cubic $x^3+x^2-n+w$ has discriminant $D = n^2+4n^3-4w-27w^2-18nw$.

$27D = 3(p-1)^2+4(p-1)^3-4(1-3p-pu)-(1-3p-pu)^2-6(p-1)(1-3p-pu)$ which after simplification gives $27D = p^2(4p-u^2) = 27p^2v^2$ and so $D = p^2v^2$

Since this is a square, the Galois group is cyclic and after computing the interpolating polynomials (or by factoring the cubic and solving the remaining quadratic), we get that if $x$ is a root then the other two are

$\frac 1v(x^2 + x(1+((n+9w)/p-v)/2) - (n+((n^2-3w)/p+v)/2))$ and its conjugate obtained by switching the sign of $v$.

Except the division by $v$ at the front, the divisions inside those coefficients all result in integers because of parity considerations, and the equations $27w \equiv -3n \equiv 1 \pmod p$.

Therefore the lattice $\Bbb Z[x_1] = \langle 1,x_1,x_1^2 \rangle$ has index $|v|$ in the lattice $\Bbb Z[x_1,x_2,x_3] = \langle 1,x_1,x_2 \rangle = \langle x_1, x_2, x_3\rangle$, and so the discriminant of this lattice is $p^2$. Since there is no unramified extension of $\Bbb Q$, it can't get any smaller and so $\Bbb Z[x_1,x_2,x_3]$ is its ring of integers and we have an integral normal basis.

Now this extension must be the class group of conductor $p$ corresponding to the unique subgroup $H$ of index $3$ in $G = (\Bbb Z/p \Bbb Z)^*$, which is trigonometric by the Kronecker Weber theorem.
Then we have another integral normal basis for the ring of integers, $\{s_K = \sum_{k \in K} \zeta_p^k \mid K \in G/H \}$

Looking at how the Galois group can act on those elements shows that we must have $x_1 = \pm s_K$ for some $K$. Finally, since $x_1+x_2+x_3 = -1 = \sum s_K$, we have $x_1 = s_K$ for some $K \in G/H$

mercio
  • 50,180
1

CW the cubics for $p < 1000,$ method of Gauss near the end of the Disquisitiones, with a leisurely modern discussion in Galois Theory by David A. Cox. Worked examples, for the primes up to 100, in REUSCHLE.

Found a table in Cohen, his "d" tends to equal $p^2,$ where mine is sometimes $4p^2$

enter image description here

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

jagy@phobeusjunior:~$ ./cubic_cyclic_gauss_loop | grep exps
  x^3 + x^2   - 2 x - 1   p  7 p.root  3 exps 6^k  d = 7^2
  x^3 + x^2   - 4 x + 1   p  13 p.root  2 exps 5^k  d = 13^2
  x^3 + x^2   - 6 x - 7   p  19 p.root  2 exps 8^k  d = 19^2
  x^3 + x^2   - 10 x - 8   p  31 p.root  3 exps 15^k  d = 2^2 * 31^2
  x^3 + x^2   - 12 x + 11   p  37 p.root  2 exps 8^k  d = 37^2
  x^3 + x^2   - 14 x + 8   p  43 p.root  3 exps 2^k  d = 2^2 * 43^2
  x^3 + x^2   - 20 x - 9   p  61 p.root  2 exps 8^k  d = 3^2 * 61^2
  x^3 + x^2   - 22 x + 5   p  67 p.root  2 exps 3^k  d = 3^2 * 67^2
  x^3 + x^2   - 24 x - 27   p  73 p.root  5 exps 7^k  d = 3^2 * 73^2
  x^3 + x^2   - 26 x + 41   p  79 p.root  3 exps 12^k  d = 79^2
  x^3 + x^2   - 32 x - 79   p  97 p.root  5 exps 19^k  d = 97^2
  x^3 + x^2   - 34 x - 61   p  103 p.root  5 exps 3^k  d = 3^2 * 103^2
  x^3 + x^2   - 36 x - 4   p  109 p.root  6 exps 2^k  d = 2^4 * 109^2
  x^3 + x^2   - 42 x + 80   p  127 p.root  3 exps 5^k  d = 2^2 * 127^2
  x^3 + x^2   - 46 x + 103   p  139 p.root  2 exps 8^k  d = 139^2
  x^3 + x^2   - 50 x - 123   p  151 p.root  6 exps 3^k  d = 3^2 * 151^2
  x^3 + x^2   - 52 x + 64   p  157 p.root  5 exps 2^k  d = 2^4 * 157^2
  x^3 + x^2   - 54 x - 169   p  163 p.root  2 exps 5^k  d = 163^2
  x^3 + x^2   - 60 x - 67   p  181 p.root  2 exps 6^k  d = 5^2 * 181^2
  x^3 + x^2   - 64 x + 143   p  193 p.root  5 exps 11^k  d = 3^2 * 193^2
  x^3 + x^2   - 66 x + 59   p  199 p.root  3 exps 12^k  d = 5^2 * 199^2
  x^3 + x^2   - 70 x - 125   p  211 p.root  2 exps 8^k  d = 5^2 * 211^2
  x^3 + x^2   - 74 x - 256   p  223 p.root  3 exps 13^k  d = 2^2 * 223^2
  x^3 + x^2   - 76 x - 212   p  229 p.root  6 exps 2^k  d = 2^4 * 229^2
  x^3 + x^2   - 80 x + 125   p  241 p.root  7 exps 17^k  d = 5^2 * 241^2
  x^3 + x^2   - 90 x + 261   p  271 p.root  6 exps 24^k  d = 3^2 * 271^2
  x^3 + x^2   - 92 x + 236   p  277 p.root  5 exps 2^k  d = 2^4 * 277^2
  x^3 + x^2   - 94 x + 304   p  283 p.root  3 exps 2^k  d = 2^2 * 283^2
  x^3 + x^2   - 102 x - 216   p  307 p.root  5 exps 2^k  d = 2^2 * 3^2 * 307^2
  x^3 + x^2   - 104 x + 371   p  313 p.root  10 exps 7^k  d = 313^2
  x^3 + x^2   - 110 x - 49   p  331 p.root  3 exps 7^k  d = 7^2 * 331^2
  x^3 + x^2   - 112 x + 25   p  337 p.root  10 exps 5^k  d = 7^2 * 337^2
  x^3 + x^2   - 116 x - 517   p  349 p.root  2 exps 6^k  d = 349^2
  x^3 + x^2   - 122 x + 435   p  367 p.root  6 exps 3^k  d = 3^2 * 367^2
  x^3 + x^2   - 124 x - 221   p  373 p.root  2 exps 8^k  d = 7^2 * 373^2
  x^3 + x^2   - 126 x + 365   p  379 p.root  2 exps 8^k  d = 5^2 * 379^2
  x^3 + x^2   - 132 x - 544   p  397 p.root  5 exps 15^k  d = 2^4 * 397^2
  x^3 + x^2   - 136 x - 515   p  409 p.root  21 exps 11^k  d = 5^2 * 409^2
  x^3 + x^2   - 140 x - 343   p  421 p.root  2 exps 8^k  d = 7^2 * 421^2
  x^3 + x^2   - 144 x - 16   p  433 p.root  5 exps 42^k  d = 2^6 * 433^2
  x^3 + x^2   - 146 x - 504   p  439 p.root  15 exps 3^k  d = 2^2 * 3^2 * 439^2
  x^3 + x^2   - 152 x - 220   p  457 p.root  13 exps 5^k  d = 2^6 * 457^2
  x^3 + x^2   - 154 x + 343   p  463 p.root  3 exps 7^k  d = 7^2 * 463^2
  x^3 + x^2   - 162 x - 505   p  487 p.root  3 exps 7^k  d = 7^2 * 487^2
  x^3 + x^2   - 166 x + 536   p  499 p.root  7 exps 2^k  d = 2^2 * 3^2 * 499^2
  x^3 + x^2   - 174 x - 891   p  523 p.root  2 exps 8^k  d = 3^2 * 523^2
  x^3 + x^2   - 180 x + 521   p  541 p.root  2 exps 8^k  d = 7^2 * 541^2
  x^3 + x^2   - 182 x - 81   p  547 p.root  2 exps 8^k  d = 3^4 * 547^2
  x^3 + x^2   - 190 x - 719   p  571 p.root  3 exps 7^k  d = 7^2 * 571^2
  x^3 + x^2   - 192 x + 171   p  577 p.root  5 exps 14^k  d = 3^4 * 577^2
  x^3 + x^2   - 200 x + 512   p  601 p.root  7 exps 31^k  d = 2^6 * 601^2
  x^3 + x^2   - 202 x - 1169   p  607 p.root  3 exps 6^k  d = 607^2
  x^3 + x^2   - 204 x + 999   p  613 p.root  2 exps 8^k  d = 3^2 * 613^2
  x^3 + x^2   - 206 x + 321   p  619 p.root  2 exps 3^k  d = 3^4 * 619^2
  x^3 + x^2   - 210 x - 1075   p  631 p.root  3 exps 27^k  d = 5^2 * 631^2
  x^3 + x^2   - 214 x - 1024   p  643 p.root  11 exps 2^k  d = 2^2 * 3^2 * 643^2
  x^3 + x^2   - 220 x - 1273   p  661 p.root  2 exps 8^k  d = 3^2 * 661^2
  x^3 + x^2   - 224 x - 997   p  673 p.root  5 exps 10^k  d = 7^2 * 673^2
  x^3 + x^2   - 230 x + 128   p  691 p.root  3 exps 2^k  d = 2^2 * 5^2 * 691^2
  x^3 + x^2   - 236 x + 1313   p  709 p.root  2 exps 8^k  d = 709^2
  x^3 + x^2   - 242 x + 1104   p  727 p.root  5 exps 3^k  d = 2^2 * 3^2 * 727^2
  x^3 + x^2   - 244 x + 1276   p  733 p.root  6 exps 2^k  d = 2^4 * 733^2
  x^3 + x^2   - 246 x - 520   p  739 p.root  3 exps 2^k  d = 2^2 * 5^2 * 739^2
  x^3 + x^2   - 250 x + 1057   p  751 p.root  3 exps 6^k  d = 7^2 * 751^2
  x^3 + x^2   - 252 x + 729   p  757 p.root  2 exps 8^k  d = 3^4 * 757^2
  x^3 + x^2   - 256 x - 1481   p  769 p.root  11 exps 7^k  d = 5^2 * 769^2
  x^3 + x^2   - 262 x - 991   p  787 p.root  2 exps 3^k  d = 3^4 * 787^2
  x^3 + x^2   - 270 x + 1592   p  811 p.root  3 exps 2^k  d = 2^2 * 811^2
  x^3 + x^2   - 274 x + 61   p  823 p.root  3 exps 5^k  d = 11^2 * 823^2
  x^3 + x^2   - 276 x - 307   p  829 p.root  2 exps 7^k  d = 11^2 * 829^2
  x^3 + x^2   - 284 x + 1011   p  853 p.root  2 exps 5^k  d = 3^4 * 853^2
  x^3 + x^2   - 286 x - 509   p  859 p.root  2 exps 8^k  d = 11^2 * 859^2
  x^3 + x^2   - 292 x + 1819   p  877 p.root  2 exps 6^k  d = 877^2
  x^3 + x^2   - 294 x + 1439   p  883 p.root  2 exps 8^k  d = 7^2 * 883^2
  x^3 + x^2   - 302 x - 739   p  907 p.root  2 exps 8^k  d = 11^2 * 907^2
  x^3 + x^2   - 306 x - 1872   p  919 p.root  7 exps 3^k  d = 2^2 * 3^2 * 919^2
  x^3 + x^2   - 312 x - 2221   p  937 p.root  5 exps 15^k  d = 937^2
  x^3 + x^2   - 322 x + 1361   p  967 p.root  5 exps 3^k  d = 3^4 * 967^2
  x^3 + x^2   - 330 x - 2349   p  991 p.root  6 exps 3^k  d = 3^2 * 991^2
  x^3 + x^2   - 332 x - 480   p  997 p.root  7 exps 2^k  d = 2^4 * 3^2 * 997^2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Will Jagy
  • 139,541