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