Why is there no solution with radicals for the quintic equation $x^5-x+1=0$?

Question

I'm looking for solution for this equation $$ x^5-x+1=0. $$ I know that there is no solution with radicals. But, I can not find possible solutions (in MSE or internet resource).

I know Abel-Ruffini theorem. But, this is so hard for general form.

Question: Can you please show me in this particular case why there is not solution with radicals?

Answer 1

If you are familiar with the use of Galois theory, then this particular polynomial can be handled as follows. Let $G$ be the Galois group of this polynomial.

1. It is irreducible over $\Bbb{Z}$ (hence over $\Bbb{Q}$) because it is irreducible over $\Bbb{Z}_5$. See this thread for many ways of seeing that. Consequently we can view $G$ as a transitive subgroup of $S_5$. In particular $G$ contains an element $\tau$ of order five which obviously must be a $5$-cycle.
2. Modulo two it factors as $$x^5-x+1\equiv(x^2+x+1)(x^3+x^2+1)$$ a product of an irreducible quadratic and an irreducible cubic. By Dedekind's theorem the group $G$ thus contains an element $\sigma$ with cycle type $(2,3)$.
3. The permutation $\sigma$ is odd, and it has order six. Therefore $|G|$ is divisible by $30$, and $G$ is not subgroup of $A_5$. It follows (from a census of subgroups of $S_5$) that $G$ must be all of $S_5$.
4. The group $G$ is not solvable, so by one of the main results of Galois theory the zeros of your polynomial cannot come from a field gotten as a root tower extension of $\Bbb{Q}$.

Answer 2

There are solutions in terms of hypergeometric functions: for example,

$${\mbox{$_4$F$_3$}(1/5,2/5,3/5,4/5;\,1/2,3/4,5/4;\,{3125}/{256})}$$

See e.g. this article.