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I have heard from various sources that the typical arithmetic operations (addition, subtraction, multiplication, division, rational exponentiation) are not sufficient to express in general the roots of a quintic polynomial. This is due to something along the lines of their inability to "express the necessary symmetries."

Would it be possible to introduce new arithmetic operations that allow do "express the necessary symmetries" and therefore allow a quintic formula to be written?

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Trivially, yes. –  Paul Draper Apr 20 at 3:38
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@DavidZhang Apart from nontriviality, yes. Hypergeometric functions and thetanulls. (latter can be found here) –  Balarka Sen Apr 22 at 15:42

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up vote 20 down vote accepted

Yes, there is such an operation called a Bring radical: http://en.wikipedia.org/wiki/Bring_radical

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Interesting. How does this Bring radical generalize to higher-order polynomials? If the inverse function of $f(x) = x^5 + x$ allows us to solve quintics, can we use the inverse function of $f(x) = x^6 + x$ or something similar to solve sextics? –  David Zhang Apr 20 at 6:33
    
@DavidZhang Bring radical generalized to higher polynomials alright, through something we call Glasser's method, but a general sextic (provably) cannot be transformed into the trinomial $x^6 + x + \alpha = 0$. The best we can do is to transform a general sextic into $x^6 + x^2 + \alpha x + \beta = 0$ which can be solved through double-variable hypergeometric functions. –  Balarka Sen Apr 21 at 19:18

You can invent your own notation: say $r(a,b,c,d,e;n)$ for the $n^\mathrm{th}$ smallest root of the quintic with coefficients $a, b, c, d, e$. You may object that this is kind of a cop-out, and you'd be right, but it's also basically what we do with quadratics. We get excited when we find out that we can "solve" $x^2 = 2$ by writing $x = \sqrt 2$, but all we've really done is say "the solution to $x^2 = 2$ is a number that, when squared, gives $2$". Which is not so insightful.

What's interesting about $\sqrt a$ is that once we can "solve" equations like $x^2 - a = 0$ we find that we don't need anything new to solve $ax^2 + bx + c = 0$ – we can transform all such equations into the simpler form. Moreover, we can develop methods of reasoning with these solutions, so that e.g. $\sqrt{x}\sqrt{y} = \sqrt{xy}$ and so forth (but note that there isn't really a great deal we can say about $\sqrt 2 + \sqrt 3$, for example). We don't so much say "these are the solutions of this equation" but "this is a relationship between the solutions of this equation and this other one", and that is something that takes genuine mathematical work and insight.

Anyway, it sounds like Julian's answer is pretty much what you wanted, but I thought I'd provide some additional colour.

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