# Could there be a special-function counterexample to Schanuel's conjecture?

It is not too hard to show that if Schanuel's conjecture is true, then the only algebraic numbers admitting a "closed-form expression" (as defined precisely in this paper) involving $e$, $\pi$, and other exponential-logarithmic constants are the ones solvable in radicals.

While reading Ken Ono's entertaining book My Search for Ramanujan recently, I was struck by the fact that some of Ramanujan's miraculous discoveries yield seemingly "transcendental expressions" for algebraic numbers. This leads to my question: Could one exploit special-function theory to construct an explicit closed-form expression for an algebraic number that is not solvable in radicals?

I expect the answer to be no, since I expect Schanuel's conjecture to be true. Still, even if that is the case, I wonder if there is any way to prove a precise theorem along these lines, that all closed-form expressions constructed in a certain way must be solvable in radicals if they are algebraic. Unfortunately, I am not familiar enough with special function theory to even tell if this question makes sense, but I was hoping some MO reader might be able to help.

EDIT: As an illustration, here's one of Ramanujan's results, reproduced from Douglas Hofstadter's book Gödel, Escher, Bach:

$${e^{-2\pi/\sqrt 5}\over\displaystyle \strut 1+{e^{-2\pi\sqrt 5}\over \displaystyle \strut 1+{e^{-4\pi\sqrt 5}\over \displaystyle \strut 1+{e^{-6\pi\sqrt 5}\over \displaystyle \strut {\ \atop 1+\cdots}}}}} = {{\sqrt 5 \over \displaystyle \strut 1+\root 5 \of {5^{3/4} \biggl({\sqrt 5 - 1 \over 2} \biggr)^{\! 5/2}\!\! -1}} -{\sqrt 5 + 1 \over 2}}$$ This doesn't answer my question directly because the expression on the left is an infinite continued fraction, which isn't a closed-form expression in my sense, but it does make me wonder whether there could be an identity of this sort where the "$\dots$" on the left-hand side can be replaced by a terminating expression.

Yes, some special cases of the hypergeometric function give roots of polynomial equations whose Galois groups are not solvable. The simplest examples are the solutions of $y(1-y)^t = x$ for rational $t$. Here is one elementary proof of this formula (which can also be obtained by Lagrange inversion); here is a simpler one that requires you to already have surmised the formula some other way (e.g. experimentally).
• It is known that every quintic equation can be reduced (perhaps with extraction of a few roots) to a value of such a hypergeometric function. There are also modular functions that do the same trick via the icosahedral cover $X(5) \to X(1)$, though you have to evaluate both a modular function and the inverse function of one. – Noam D. Elkies Sep 15 '16 at 0:12