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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?

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  • $\begingroup$ Even wolframalpha doesn't know the exact forms for the solutions, only approximations. $\endgroup$ – JMoravitz Nov 6 '17 at 6:15
  • $\begingroup$ I know wolfram alpha.. $\endgroup$ – Math Nov 6 '17 at 6:17
  • $\begingroup$ To see why there are no solutions with radicals, see this answer (for a slightly different polynomial). $\endgroup$ – Mark Nov 6 '17 at 6:18
  • $\begingroup$ 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. $\endgroup$ – ultrainstinct Nov 6 '17 at 6:19
  • $\begingroup$ That is strange: Bring radical has not been mentioned yet. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 13:12
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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.

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  • $\begingroup$ maybe, after 1000 years, can radical solutions be found? $\endgroup$ – Math Nov 6 '17 at 6:36
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    $\begingroup$ No: theorems never get repealed. $\endgroup$ – Robert Israel Nov 6 '17 at 7:35
  • $\begingroup$ Teacher,Can you explain me, is this function express all $5$ roots? Or 1$ root? $\endgroup$ – Math Apr 9 '18 at 21:26
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    $\begingroup$ This is one root. There are ways to express the others. $\endgroup$ – Robert Israel Apr 10 '18 at 22:57
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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 $\Bbb{Q}$.
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  • $\begingroup$ 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? $\endgroup$ – Arthur Nov 6 '17 at 8:37
  • $\begingroup$ @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 :-/ $\endgroup$ – Jyrki Lahtonen Nov 6 '17 at 9:48
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    $\begingroup$ Wow... Thank you for amazing answer. $\endgroup$ – Math Nov 6 '17 at 11:03

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