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We know that polynomials up to fourth degree have closed solutions using radicals. And we know that starting from the quintic no polynomial will have a closed solution using radicals.

Question 1: What I want to know is, why does this happen for the fifth order polynomial? What is so special about the number 5? I don't want a prove. I am looking for an intuitive explanation.

Question 2: Is there a closed formula for higher order polynomials using other functions and operations? I heared that Felix Klein did something like that, but I could not find a closed formula for the solution.

MrYouMath
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    For question 1, it is because it is the smallest values of $n$ such that the symmetric group of degree $n$ has a non-abelian simple subgroup. That this is the case is pretty much a matter of combinatorics. – Tobias Kildetoft Apr 08 '16 at 08:14
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    Nice ... could you describe in further detail? – MrYouMath Apr 08 '16 at 08:15
  • Of which part? The first is related to Galois theory (though I think it might be doable without), and the second is a matter of counting permutations of certain types. – Tobias Kildetoft Apr 08 '16 at 08:16
  • This answer may help with the first part. http://math.stackexchange.com/a/117495/259262 – Patrick Stevens Apr 08 '16 at 08:20
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    As to your second question, there is a method of solving the quintic that was developed in the 19th Century, discussed in King's Beyond the Quartic Equation [ http://www.springer.com/us/book/9780817648367 ] ; needless to say, one resorts to non-elementary functions. This page [ https://en.wikipedia.org/wiki/Bring_radical ] may give you some idea of what you're in for... – colormegone Apr 08 '16 at 08:24
  • Rather than going directly to Galois, it may make sense to start with the Abel-Ruffini theorem since it precedes Galois and it's construction occurs with more basic tools – Phillip Hamilton Apr 08 '16 at 13:34

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If by ‘closed formula’, you mean a formula with radicals and ordinary arithmetic operations, the general answer comes from Galois theory:

One consider the Galois group of the equation, i.e. the group of automorphisms of the splitting field of the polynomial. Suppose the polynomial has degree $n$. As an automorphism of this splitting field maps a root of the equation to another root, it is a subgroup $G$ of the symmetric group $S_n$.

Now the equation is ‘solvable by radicals’ if and only if $G$ is a solvable group.

This means there is a sequence of subgroups: $$\{e\}=G_0\subset G_1\subset\dots\subset G_r=G$$ such that each $G_i$ is normal in $G_{i+1}$ and $G_{i+1}/G_i$ is abelian.

It happens that for the general polynomial of degree $n$ (i.e. with indeterminate coefficients) the Galois group is $S_n$, and that the alternating group $A_n$ is simple for $n\ge 5$, and of course, not abelian. Hence $S_n$ is not solvable.

Bernard
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