TITLE: Intuitive reasoning why are quintics unsolvable
QUESTION [52 upvotes]: I know that quintics in general are unsolvable, whereas lower-degree equations are solvable and the formal explanation is very hard. I would like to have an intuitive reasoning of why it is so, accessible to a bright high school student, or even why it should be so. I have also read somewhere that any $n$-degree equation can be depressed to the form $ax^n + bx + c$. I would also like to know why or how this happens, at least for lower degree equations. 
I know that this question might be too broad and difficult, but this is a thing that has troubled me a lot. To give some background, I recently figured out how to solve the cubic and started calculus, but quartics and above elude me. 
EDIT: It was mentioned in the comments, that not every $n$-degree equation can be depressed to the form $ax^n + bx + c$, although I recall something like this I have read, anyways, I wanted to find out the same for quintics.

REPLY [5 votes]: It is important to understand that Abel and Galois solved somewhat different problems while dealing with quintic equations. Abel always dealt with the simpler case of a general polynomial of fifth degree with symbolic coefficients where Galois handled the more general case where the coefficients involved could be numeric.
I read Abel's treatment (with some simplification done by later authors) in the wonderful book Galois Theory of Algebraic Equations by Jean Pierre Tignol. The idea behind the proof is to understand the relation between roots and coefficients with some focus on how roots are more complicated entities than the coefficients. It is easy to establish the expressions for coefficients in terms of roots (elementary symmetric functions) but doing the reverse is hard. Abel's proof for the unsolvability of a general quintic has two main ingredients :

If a root can be expressed as a radical expression in terms of coefficients (and the radicals may be nested) then the building blocks of this complicated expression (ie the innermost radicals) as well as the intermediate expressions can be expressed as rational functions of the roots using roots of unity. This is a non-trivial part of the proof but it is based on algebraic manipulation and does not deal with any permutation stuff. 
Radical expression of the coefficients (which are rational functions of the roots via first bullet point above) retain some amount of symmetry under permutation of the roots, no matter how many levels of nested radicals are involved and thus can not be used to express the roots which are totally non-symmetric. In general the process of forming radicals loses some amount of symmetry, but if the number of variables involved is $5$ or more then some amount of symmetry is always retained.

To elaborate the second point the expression $(a-b) ^{2}$ is symmetric in $a, b$ but taking a square root gives $a-b$ or $b-a$ and both of these are not symmetric in $a, b$. So taking square roots or $n$'th roots does break symmetry. But if we have a rational symmetric function say $f(a, b, c, d, e) $ of five variables and if the $n$'th root $\sqrt[n] {f(a, b, c, d, e)} $ can also be expressed as some rational function $g(a, b, c, d, e) $ then $g$ is not entirely non-symmetric ie there are some permutations of the variables which leave this expression unchanged. And this happens no matter how complicated radical expressions we generate via nesting. This fact can be demonstrated without much effort and is significantly simpler to understand compared to studying Galois theory. I have presented the details with proofs in my blog posts.