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(This post is indebted to Oscar Lanzi.)

Part I. $\color{blue}{p = 11}$

The equation,

$$\big(4\sin[3t] - \tan[t]\big)^2 = 11$$

seems to have five solutions, given by $t = \frac{2\pi\,k}{11}$ for $k = 1,2,3,4,5$, so it ranges from $t = 0.57\; \text{to}\; 2.85$. However, if you look at its plot, it intersects the $x$-axis at SIX points, the extra one being less than $0.5,$

enter image description here

As pointed out by Lanzi in an answer, this extra point in fact has a nice closed form involving the tribonacci constant $T$,

$$t = \arccos\big(T/2\big) = 0.40362481\dots$$

the real root of $T^3 = T^2+T+1$.

Part II. $\color{blue}{p = 23}$

I always wonder if things could be generalized, and Lanzi provided the equation for $p=23,$

$$\big(4\sin[2t] + 4\sin[3t] - 4\sin[5t] + 4\sin[6t] - \tan[t]\big)^2=23$$

Analogously, this seems to have eleven solutions, given by $t = \frac{2\pi\,k}{23}$ for $k = 1,2,\dots 11$, so it ranges from $t = 0.2\; \text{to}\; 3.1$. But it intersects the $x$-axis at TWELVE points, the extra one between $1.7$ and $1.8$,

enter image description here

I suspected this would involve the plastic constant $P$, the real root of $P^3 = P+1$ since $T$ is for $\mathbb{Q}\big(\sqrt{-11}\big)$ while $P$ is for $\mathbb{Q}\big(\sqrt{-23}\big)$. Using Mathematica's FindRoot command and its integer relations subroutine, its closed-form apparently is,

$$t = \arccos\big(\tfrac{1-P}2\big) = 1.73387721\dots$$

Part III. $\color{blue}{p = 31}$

Question: Anybody can find the equation for $p=31?\,$ I assume this may involve the supergolden ratio $S$, the real root of $S^3=S^2+1$ since $S$ is for $\mathbb{Q}\big(\sqrt{-31}\big)$, but that's just a guess.

Tito Piezas III
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    Done for $p=31$. Note that the polynomial degree is actually not favorable for obtaining the supergolden ratio directly. We were lucky with $p=11$ and $p=23$. – Oscar Lanzi Dec 18 '22 at 01:35
  • @OscarLanzi While $p=7$ is symmetric and has no extra solutions, it turns out the symmetric version of $p=19$ DOES have extra roots, and depend only on quartics. (The non-symmetric version depend on a $9$-deg.) Kindly see my modified addendum answer below. – Tito Piezas III Dec 19 '22 at 16:49

3 Answers3

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For $p=31$ the symmetrized sine-tangent relation is

$-\tan(\theta)+4[\sin(2\theta)+\sin(7\theta)+\sin(10\theta)-\sin(12\theta)-\sin(13\theta)+\sin(14\theta)]=(k|31)\sqrt{31}, \theta=2k\pi/31.$

For $k=1,2,3,...,15$ where the Legendre symbol $(k|31)$ is $\pm1$, this gives

$[-\tan(\theta)+4[\sin(2\theta)+\sin(7\theta)+\sin(10\theta)-\sin(12\theta)-\sin(13\theta)+\sin(14\theta)]]^2=31.$

Using appropriate trigonometric identies to express all variables in terms of $x=2\cos(\theta)$ leads to a 30th-degree polynomial in $x$. Since fifteen roots must be in the form $x=2\cos(2k\pi/31)$, the polynomial factors into two fifteenth-degree polynomials, thus:

$P_{15}(x)Q_{15}(x)=0$

where $P$ has the trigonometric roots and $Q$ has the additional coupled roots. Because $Q$ is not a cubic polynomial as was found with $p=11$ and $p=23$, the supergolden ratio fails to emerge immediately. With the 15th-degree polynomial that is actually obtained, three possibilities seem likely:

  1. The supergolden ratio is obtained because $Q$ is further reducible to give the required cubic factor.

  2. The supergolden ratio is buried in a quintic factor over $\mathbb Q[\psi]$.

  3. No extraction of the supergolden ratio is possible after all.

UPDATE: It appears, unfortunately, that possibility (3) is the one that holds.

  • Tito Piezas used Magma to show that $Q$ has no zeroes expressible in radicals, ruling out (1).

  • Inspection of computed results for $x=2\cos(\theta)$ from $+2$ to $-2$ in intervals of $0.001$ reveals 24 real roots in this range, meaning no more than three pairs of complex-conjugate roots (the possibility of roots where $x$ is real but $\theta$ is not was not considered). Possibility (2) would have required at least five such pairs, corresponding to quintic factors containing the complex conjugate roots of the supergolden ratio cubic.

Thus the supergolden ratio is not retrieved using the quadratic Gauss sum and associated sine-tangent relation with $p=31$.

However, the supergolden ratio has been obtained from a similar relation with $p=3$. See Tito's answer.

Oscar Lanzi
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  • Thanks for the edit. I checked your old $15$-deg equation and while its discriminant had the factor $31^7$, the Magma calculator says it is NOT solvable in radicals. And the $9$-deg equation for $p=19$ was also not solvable. But I DID FIND the supergolden ratio. It was in $\color{blue}{p=3}$. I will write an answer. – Tito Piezas III Dec 18 '22 at 07:16
  • The equations for $p=4n+3$, namely $p = 3,7,11,19,23,31$ have the sum $\sqrt{p}$, but I noticed for $p=4n+1$, namely $p=5,13,17,29,37$ have it as $\sqrt{a+b\sqrt{p}}$. Is this always the case? – Tito Piezas III Dec 18 '22 at 07:35
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Thanks for Lanzi for posting the equation for $p=31$. However, it turns out the supergolden ratio $S$ appeared in $p=3,$

$$\big(4\sin[t]-\tan[t]\big)^2 = 3^3$$

This equation has the unexpected and expected solutions,

$$\begin{align} t &= \arccos\Big(\tfrac{(S-1)^2}{2}\Big) = 1.46\dots\\ t &= 2\pi/3 = 2.09\dots \end{align}$$

where $S$ is the real root of $S^3=S^2+1.$ This is clearly seen in the plot below,

enter image description here

Tito Piezas III
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  • How did you get that relation? The usual Gauss sum method, applied with $p=3$, gives only $\tan(2\pi/3)=\sqrt3$, from which we get only the root $x=2\cos(2\pi/3)=1$ and periodic equivalents. – Oscar Lanzi Dec 18 '22 at 10:58
  • @OscarLanzi I reverse-engineered it by assuming a pattern for $p = 4n+3$. I made a table in the answer below for $p = 3,7,11,\color{red}{15},19,23,\color{red}{27},31$ though I am missing the ones in red. Does your method work for non-prime $p$? (I may have found something for $p=15$ with extra roots that have a closed-form.) – Tito Piezas III Dec 18 '22 at 15:37
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(This is an addendum to my answer.) It seems best to gather all the equations for small $p$ together. For $p=4n+3$,

$$\big(4\sin[t] - \tan[t]\big)^2 = \color{blue}3^3$$

$$\big(4\sin[2t] - \tan[t]\big)^2 = 7$$

$$\big(4\sin[3t] - \tan[t]\big)^2 = \color{blue}{11}$$

$$\,\big(4(\sin[6t] - \sin[12t] - \sin[8t]) - \tan[t]\big)^2=\color{blue}{19}\,$$

$$\big(4(\sin[6t] + \sin[7t] - \sin[8t])\, - \tan[t]\big)^2=19$$

$$\big(4(\sin[2t] + \sin[3t] - \sin[5t] + \sin[6t]) - \tan[t]\big)^2=\color{blue}{23}$$

$$\big(4(\sin[2t]+\sin[7t]+\sin[10t]-\sin[12t]-\sin[13t]+\sin[14t])-\tan[t]\big)^2=31$$

Solutions to these equations are $t = \frac{2\pi\,k}p$ for $k = (1,2,\dots \frac{p-1}2)$. Some remarks:

  1. For $p=7$, it intersects the $x$-axis 6 times, but is symmetric in the range $0$ to $\pi$, so involves just three roots.
  2. The cases $p=3,11,23$ have one extra solution $t$ that depend on cubic roots: the supergolden ratio, the tribonacci constant, and the plastic constant, respectively.
  3. The symmetric version of $p=19$ has two extra $t_i$ that depend on quartic roots,$$t_i = \frac12\arccos\Big(\frac{u_i}{2}\Big)$$ where the $u_i$ are the two real roots of $u^4 + u^3 - 3u^2 - 4u - 3=0$ and discriminant $d = 19^2\times23.$
  4. The non-symmetric version of $p=19$ also has extra solutions, but is determined by an equation with an unsolvable Galois group and is of $\frac{19-1}2=9$-deg.
  5. The symmetric and non-symmetric versions of $p=31$ have extra solutions determined by non-solvable equations, with the latter a $\frac{31-1}2=15$-deg.
  6. The symmetric and non-symmetric versions of $p=43$ have extra solutions determined by non-solvable equations, with the latter a $\frac{43-1}2=21$-deg.

and so on, apparently. (I used Magma to determine the Galois group and Mathematica to find prime $p=31, 43, 47$, with $p=31$ also found by Lanzi in his answer.) From this trend, it indeed would be remarkable if a sine-tangent equation for prime $p>23$ will have extra solutions that depend on radicals.

Tito Piezas III
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