Closed form solution for $x^n + x+ C = 0$

Question

Is there a closed form solution for $x^n + x+ C = 0$ where $n\in \mathbb{N}$, $C\in\mathbb{R}$?

Answer 1

The roots of the polynomial $x^5 + x + a$ can be written in terms of radicals and the Bring radical. The Bring radical is necessary and is sufficient for all quintics (via a reduction first to Bring-Jerrard form).

The Bring radical is insufficient for higher degrees. For instance, the Kampé de Fériet function is used to solve the sextic (and almost no one would call an expression containing that function "closed").

I recall (and would entertain a reference in comments or a counterexample, since I don't have one handy) that, in degree $\geq 5$, solvable polynomials are isolated in the space of coefficients, so even if there are an $n \geq 5$ and $C$ which give a polynomial with roots in terms of radicals, there is an interval $(C - \epsilon, C + \epsilon)$ such that if $C'$ is in that interval and $C' \neq C$, then $x^n + x + C'$ does not have roots expressible in radicals. (That is, a microscopic change in $C$ breaks the property of having roots expressible in radicals, no matter how small the change.)

Answer 2

. Let $x=\zeta_{n-1}\bar x$, $\bar C=-\zeta_{n-1}C$, and multiply both sides by $-\zeta_{n-1}$, where $(\zeta_m)^m=-1$, to get

$$\bar x^n-\bar x-\bar C=0$$

By the Lagrange inversion theorem, a series solution is given by

$$\bar x=\sum_{k=0}^\infty\binom{nk}k\frac{\bar C^{(n-1)k+1}}{(n-1)k+1}$$

In terms of the generalized hypergeometric function this is

$$\bar x=\bar C\cdot{}_nF_{n-1}\left(\frac1n,\frac2n,\dots,\frac nn;\frac2{n-1},\frac3{n-1},\dots,\frac n{n-1};\frac{n^n}{(n-1)^{n-1}}\bar C^{n-1}\right)$$

and hence the original solution is given by

$$x=-\zeta_{n-1}^2\cdot{}_nF_{n-1}\left(\frac1n,\frac2n,\dots,\frac nn;\frac2{n-1},\frac3{n-1},\dots,\frac n{n-1};\frac{n^n}{(1-n)^{n-1}}C^{n-1}\right)$$

As noted in the comments and other answers, this only reduces down to a reasonable closed-form in general for $n<5$. For $n=5$, the Bring radical is needed, and from there it just gets nastier.

Answer 3

If you consider the hypergeometric function and generalized hypergeometric function ${_pF_q}$ as closed-forms, then there are closed-form solutions to $x^n+x-\beta = 0\,$ for any integer $n>2$.

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I. n = 5

$$z^5+z-\beta = 0$$

A root $z$ is,

$$z = \color{blue}\beta\,{_4F_3}\Big(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,-\frac{5^5}{4^4}\beta^4\Big)$$

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II. n = 6

$$z^6+z-\beta = 0$$

A root $z$ is,

$$z = \color{blue}\beta\,{_5F_4}\Big(\frac16,\frac26,\frac36,\frac46,\frac56;\,\frac25,\frac35,\frac45,\frac65;\,-\frac{6^6}{5^5}\beta^5\Big)$$

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And so on.

Answer 4

$$x^n+x+C=0$$

This equation is a trinomial equation. See [Szabó 2010].
$\ $

$$x+x^n+C=0$$ $$-\frac{x}{C}-\frac{x^n}{C}-1=0$$ $x\to -Ct$: $$t-(-1)^nC^{n-1}t^n-1=0$$

Now the equation is in the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $\ _1\Psi_1$ can be obtained therefore.
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[Szabó 2010] Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104

[Belkić 2019] Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106