The unsolvability of a general quintic equation in terms of the basic arithmetic operations and $n$th roots (i.e. the Abel–Ruffini theorem) is considered a major result in the mathematical canon. I have recently become confused as to why this is the case.

The formula $z=\frac{-b+\sqrt{b^2-4ac}}{2a}$ expresses the solutions to the quadratic equation $az^2+bz+c = 0$ in terms of the inverse of an analytic function $z \mapsto z^2$. We have simply turned the problem of inverting one analytic function, $z \mapsto az^2+bz$, into the problem of inverting another analytic function, $z \mapsto z^2$. Therefore, all the power of the quadratic equation lies in how it solves *any* quadratic equation by inverting a *single* analytic function, $z \mapsto z^2$.

Similarly, Cardano's formula solves *any* cubic equation by inverting a two analytic functions, $z \mapsto z^2$ and $z \mapsto z^3$. Interestingly, you can also solve a cubic by inverting only one analytic function, for example $z \mapsto \sin z$.

And crucially, you can also do this for quintic equations, by inverting $z \mapsto z^k, k \leq 4$ and $z \mapsto z^5+z$.

One possible statement of the Abel–Ruffini theorem is that it is impossible to solve a general quintic equation by exclusively inverting functions of the form $z \mapsto z^k$ for $k \in \mathbb{N}$. But why would we only be interested in solutions that invert analytic functions of that form? In simplier terms, **what's so special about radicals that makes solutions in terms of them so desirable?** I can't see an argument that such inverses are intuitively straightforward: they often produce answers that purely formal (e.g. $\sqrt{2}$ doesn't have a simpler defintion than *the positive inverse of the squaring function at $2$*).

To me, it seems that the more natural question is, for $n \in \mathbb{N}$,

Is there a finite set of analytic functions such that the solutions to any degree $n$ polynomial may be expressed in terms of the inverse of these analytic functions?

I know very little about the status of this question (exept that it holds for some small values of $n$). **Any information on what is known about this question would also be of interest.**

notpreferred because of the corresponding field extensions. The idea of a radical formula just came out of the discovery of root formulas in low degree. That there is a translation of the concept into the language of field extensions is simply part of the process of explaining how to turn the existence of a radical formula into the existence of a certain structure in a Galois group using today’s interpretation of Galois theory. Nobody was thinking about field extensions 200 years ago. $\endgroup$answerit was pretty mind-expanding. The fact that Galois theory is a broader context that is useful elsewhere was just icing in the cake. I've moved onto harder (mathematical) drugs by now, but surely "answers a classical question that many students will have at least implicitly thought about" is a good reason to teach something... $\endgroup$36more comments