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

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?

• Even wolframalpha doesn't know the exact forms for the solutions, only approximations. – JMoravitz Nov 6 '17 at 6:15
• I know wolfram alpha.. – Math Nov 6 '17 at 6:17
• To see why there are no solutions with radicals, see this answer (for a slightly different polynomial). – Mark Nov 6 '17 at 6:18
• In general, if the polynomial has rational roots we can apply en.wikipedia.org/wiki/Rational_root_theorem. But other than that, you would need numerical methods to find them. – ultrainstinct Nov 6 '17 at 6:19
• That is strange: Bring radical has not been mentioned yet. – Jack D'Aurizio Nov 6 '17 at 13:12

$${\mbox{_4F_3}(1/5,2/5,3/5,4/5;\,1/2,3/4,5/4;\,{3125}/{256})}$$
• Teacher,Can you explain me, is this function express all $5$ roots? Or 1$root? – Math Apr 9 '18 at 21:26 • This is one root. There are ways to express the others. – Robert Israel Apr 10 '18 at 22:57 If you are familiar with the use of Galois theory, then this particular polynomial can be handled as follows. Let$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$\Bbb{Q}$. • Dedekind's theorem with modulo 5 seems to prove that there is a$5$-cycle in$G$. However, I don't see how it follows from$G$being transitive. Is it a general fact that all transitive subgroups of$S_n$contains an$n$-cycle, or is there something else at work here? – Arthur Nov 6 '17 at 8:37 • @Arthur. No, not all transitive subgroups of$S_n$contain an$n$-cycle. The smallest example is that copy of (a transitive) Klein four as a subgroup of$A_4$. But, when$n$is a prime,$n=p$, then a transitive subgroup$G$of$S_p$does contain a$p$-cycle. This is because by orbit-stabilizer we have$p\mid |G|$, so by Cauchy there is an element of order$p$in$G$. But such an element is necessarily a$p\$-cycle. Sorry about leaving the answer a bit too sketchy at that point :-/ – Jyrki Lahtonen Nov 6 '17 at 9:48