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.




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**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.




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**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.




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**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.)




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**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.




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**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?



A method to generate solvable equations of degrees $p = 7, 13, 19, 31, 37,\dots$ using only cubics
The following can explain the shape of the expression and also why you always get pairs (why the minimal polynomials for the two elements in a pair differ by few coefficients, I don't know).


Let $p$ be a prime congruent to $1$ mod $6$, and consider the rational function $f(X):=g(X^p)$, where $g(X):=\frac{X^3-3X+1}{X-X^2}$ (note that setting $t:=g(X)$ and clearing denominators gives exactly your generic cubic, except that I like $t$ better than $n$ as a parameter). Then $X\mapsto f(X)$ yields a cover $\mathbb{P}^1\to \mathbb{P}^1$, and we can try to compute its monodromy group (i.e., the Galois group of $f(X)-t=0$ over $\mathbb{C}(t)$ - and, as a matter of fact, still over $\mathbb{Q}(\zeta\_p)(t)$). Since $Gal(g(X)-t/\mathbb{C}(t)) = C\_3$ and $Gal(X^p-t/\mathbb{C}(t))=C\_p$, this monodromy group will be a subgroup of $C\_p\wr C\_3 (=(C\_p\times C\_p\times C\_p)\rtimes C\_3) \le S\_{3p}$. We can be more explicit: IF $a,b,c$ denote the three roots of $g(X)-t=0$, and $\alpha:=a^{1/p}$, $\beta:=b^{1/p}$, $\gamma:=c^{1/p}$, then the roots of $f(X)-t=0$ fall into three blocks $\Delta\_1, \Delta\_2, \Delta\_3$, with $\Delta\_1:=\{\alpha, \zeta\alpha, \dots, \zeta^{p-1}\alpha\}$ (and analogously for $\Delta\_2,\Delta\_3$).


The inertia groups at the branch points of the cover $X\mapsto f(X)$ will now give more precise information about the Galois group. It's easy to see that $X\mapsto g(X)$ has exactly two (non-rational) branch points (namely the roots of your discriminant $D$), whereas $X\mapsto X^p$ has of course exactly the branch points $0$ and $\infty$, and these two get mapped to the same point $\infty$ under $X\mapsto g(X)$. So in total, the composition has exactly three branch points: the two irrational ones (whose inertia group generators are of order $3$ (permuting $a$, $b$, $c$ cyclically and thus permuting the three blocks $\Delta\_i$ in the same way), and $t=\infty$ at which the inertia group is a double-$p$-cycle inside the block kernel (since unramified under $X\mapsto g(X)$, but with two totally ramified preimages on the "upper" level). The permutation group generated by these elements can then be identified as $G:=K\rtimes C\_3$, where $K:=\{(x\_1,x\_2,x\_3)\mid x\_1,x\_2,x\_3\in \mathbb{F}\_p; \sum x\_i=0\}$ is the ``augmentation ideal". This group $G$ has two normal subgroups $N\_1$ and $N\_2$ isomorphic to $C\_p$, both with quotient group $G/N\_i\cong C\_p\rtimes C\_3$: namely, let $y\_1,y\_2$ be the two primitive third roots of unity in $\mathbb{F}\_p$ and let $N\_i:=\{(x,y\_i x, y\_i^2 x): x \in \mathbb{F}\_p\}\trianglelefteq K$.
We want to find an expression in $\alpha,\beta,\gamma$ which is fixed by $N\_i$, and claim that this can be done in the form $\alpha^j\beta^k$ for $j,k \in \{1,\dots, p-1\}$ suitable, so that the minimal polynomial of $\alpha^j\beta^k (=a^jb^k)^{1/p}$ over $\mathbb{C}(t)$ will have Galois group $G/N\_i\cong C\_p\rtimes C\_3$ (the remaining symmetrization $\alpha^j\beta^k + \beta^j\gamma^k + \gamma^j\alpha^k$ the only drops the degree by a factor $3$, reaching a degree-$p$ polynomial, but still with the same splitting field; the existence of such a polynomial is clear anyway, since $C\_p\rtimes C\_3 \le S\_p$, so from the theoretical point of view this last step is not so important anymore).
So let $\sigma\_i:=(1,y\_i,y\_i^2)\in N\_i$ be a generator of $N\_i$. This acts quite explicitly on the three blocks $\Delta\_1,\Delta\_2,\Delta\_3$ by
$\sigma\_i(\alpha) = \zeta\_p \alpha$, $\sigma\_i(\beta) = \zeta\_p^{y\_i}\beta$, $\sigma\_i(\gamma) = \zeta\_p^{y\_i^2}\gamma$.


Hence, $\sigma\_i(\alpha^j\beta^k) = \zeta\_p^{j+k\cdot y\_i} \alpha^j\beta\_k$, and we're laughing if we can get $j+k\cdot y\_i$ divisible by $p$. There's obviously some freedom here in choosing $j$. E.g., for $p=7$, the third roots of unity in $\mathbb{F}\_7$ are $y\_1=2$ and $y\_2=4$; now if you choose $j=3$ with $y\_2=4$, you get $k=1$, which after the symmetrization yields your first solution $x\_7$; whereas $j=1$ with $y\_1=2$ gives $k=3$, and that, after symmetrization is your second solution $x\_7'$.


Just to check one more case, for $p=37$, the third roots of unity are $10$ and $26$. Choosing $j=7$ with $y\_1=10$ gives $k=3$.
You could of course choose other exponents $j$ (whether choosing to increase $j$ by $1$ for each new prime is actually a natural choice, I don't know; it happens to give small $k$ for the first few terms, but if I made no mistake, choosing $j=8$ for $p=43$ should give you $k=13$; instead always choosing $j=1$ would give the rather systematically-looking $k=y\_i+1$!), although these would of course not generate new field extensions.


