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I. Reduced equations

One can eliminate 3 terms from the general quintic, sextic, septic, and octic using a Bring-Jerrard transformation to get the reduced forms in radicals,

$$x^5+(x+p) = 0$$ $$x^6+(x+p)(x+q) = 0$$ $$x^7+(x+p)(x+q)(x+r) = 0$$ $$x^8+(x+p)(x+q)(x+r)(x+s) = 0$$

Note: It is assumed the constant term $pqrs \neq 0$. Incidentally, Hilbert, using a geometric approach involving 27 lines, managed to eliminate four terms from the general nonic reducing it to,

$$x^9+(x+p)(x+q)(x+r)(x+s) = 0$$

though it seems one may have to solve a deg-5 or deg-6, hence may not be in radicals. (See Section 4 of Sutherland's 2021 paper.)

II. Solutions to the above

The series solution to the reduced degree $m=5$ is well-known,

$$x = -\sum_{k=0}^\infty(-1)^k\frac{(5k)!}{k!(4k+1)!}\;p^{4k+1}$$

which can be extended via analytic continuation using the generalized hypergeometric function ${_4F_3}.$ In these posts here and here, user Tyma Gaidash also found series solutions to the reduced sextic, septic, and octic given above as,

$$x_k=\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}6 \frac{q^\frac n6\left(\frac n6\right)!}{p^{\frac{5n}6-1}\big(1-\frac{5n}6\big)!\,n!}\,{_2 F_1}\left(1-n,-\frac n6;2-\frac{5n}6;\color{blue}{\frac pq}\right)$$

$$x_k =\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}7 \frac{(qr)^\frac n7\left(\frac n7\right)!}{p^{\frac{6n}7-1}\big(1-\frac{6n}7\big)!\,n!}\, \operatorname{F}_1\left(1-n;-\frac n7,-\frac n7;2-\frac{6n}7;\color{blue}{\frac pq,\frac pr}\right)$$

$$x_k=\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}8\frac{(qrs)^\frac n8\left(\frac n8\right)!}{p^{\frac{7n}8-1}\left(1-\frac{7n}8\right)!\,n!}\, F_D^{(3)}\left(1-n;-\frac n8,-\frac n8,-\frac n8;2-\frac{7n}8;\color{blue}{\frac pq,\frac pr,\frac ps}\right)$$

for all roots $x_k$ with $k = 0,1,\dots m-1,$ using the hypergeometric function ${_2F_1},$ Appell function $\text{F}_1,$ and Lauricella function $F_D^{(3)}$, respectively.

III. Hilbert's 13th

Hilbert's 13th Problem asked if the reduced septic's roots, considered as a function of the three variables $(p,q,r)$, can be expressed as the composition of a finite number of two-variable functions. Tyma's solution for the septic does use the two-variable Appell function $\text{F}_1,$ though one must sum an infinite number of them.

IV. Questions

  1. Why does Hilbert start with the septic? Is it known that the reduced sextic's roots, as a function of the two parameters $(p,q)$ can be expressed as a composition of a finite number of one-variable functions?
  2. And via analytic continuation, are Tyma's series the complete solution for the reduced sextic, septic, and octic for arbitrary $(p,q,r,s)$ as long as its constant term $pqrs \neq 0$?
Tito Piezas III
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  • The series do not converge seemingly if one of the parameters has absolute value less than one. – Tyma Gaidash Jul 04 '23 at 11:59
  • @TymaGaidash It says in this article that Appell $F_1(a,b_1,b_2;c;\beta_1,\beta_2)$ is defined for $|\beta_1|<1,,|\beta_2|<1$, but, for other $(\beta_1,\beta_2)$, can be extended by analytic continuation and gave another formula. So there should be a way to make it converge. – Tito Piezas III Jul 04 '23 at 12:26
  • Related MO question. – Tito Piezas III Jul 04 '23 at 13:30
  • Possibly related: https://mathoverflow.net/questions/422922/can-a-general-quintic-be-solved-using-inverse-beta-regularized-function – Anixx Aug 10 '23 at 05:08
  • Can you please explain how the infinite-series solutions are related to the Hilbert problem which is about closed-form solutions? – Anixx Aug 10 '23 at 05:16
  • @Anixx More precisely, Hilbert's 13th asks if the reduced septic $x^7+ax^3+bx^2+cx+1$ with three parameters can be solved using a finite composition of a function of two variables. It is well-known that the reduced sextic can solved by the two-variable Kampe de Feriet function. So it seems surprising that the septic can be solved by a two-variable function (the Appell F1) as well. The non-radical solution to the Bring quintic is a series solution called the Bring radical. Therefore, series solutions are not automatically excluded in the toolkit of solving equations, you know. – Tito Piezas III Aug 10 '23 at 05:53
  • Each term in the septic expansion above is a function of 3variables, p, q and r (not counting the root index k). – Anixx Aug 10 '23 at 08:56
  • @Anixx: Oh, you mean,the factors $(p,q,r)$ outside the Appell function? I believe we just express it as powers of reciprocals of $\beta_1 = \frac{p}{q}, \beta_2 = \frac{p}{r}$, hence $\frac{q}{p}$ and $\frac{r}{p}$. – Tito Piezas III Aug 10 '23 at 09:37
  • @TitoPiezasIII The coefficients in each term outside the $F$ functions also have p's and q's. – Anixx Aug 10 '23 at 10:52
  • @Anixx Yes, there may be transformations to have the $F$ functions absorb those factors. I will notify Tyma if he has suggestions. – Tito Piezas III Aug 10 '23 at 11:24

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