I am aware this is a pretty big topic, but the attempts at layman's explanations I have seen either barely provide commentary on the formal proofs, or fail to provide an explanation (e.g "it gets too complex" does not really say anything)
Is there a good intuitive explanation as to why we fail to obtain a general solution for a 5th+ degree polynomial, and why this happens at the 5th degree and not below or above?
See this detailed sketch of Arnold's proof of Abel-Ruffini theorem here: web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.pdf. It is intended for a layman who can handle some basic group theory (and a bit of topology). It still takes about 5 pages.
Responding to Dmitry Ezhov's comment above: the "finite combination" is necessary, since with an infinite number of operations one can solve an arbitrary quintic. For example, consider $x^5 - x - 1$. Galois theory tells us the roots of this cannot be expressed in terms of a finite combination of radicals and field operations, but using infinitely many it's not too bad: if $x^5 - x - 1 = 0$, then $x^5 = x+1$, so $x = \sqrt[5]{1+x}$. Plugging this back into itself and iterating yields a solution $x = \sqrt[5]{1+\sqrt[5]{1+\cdots}}$ to the original quintic.
Formula for roots of equation $\displaystyle z^m-az^n-1=0$ with definite integration:
$\displaystyle z_j=e^{2j\pi i/m}+\frac{1}{2\pi i}\left(e^{(2j+1)\pi i/m}\int_0^\infty log\left(1+a\frac{t^n}{1+t^m}e^{(2j+1)\pi in/m}\right)dt \\- e^{(2j-1)\pi i/m}\int_0^\infty log\left(1+a\frac{t^n}{1+t^m}e^{(2j-1)\pi in/m}\right)dt\right)$
where natural $m>n>0$, $j=0,1,...m-1$ and $a$ is natural.
Based on paper Лахтинъ, “Выраженiе корней трехчленнаго алгебраическаго уравненiя посредствомъ опредѣленныхъ интеграловъ” (1890).
Example calculation in pari/gp:
a= 7;
m= 5; n= 2;
print("Quintic: z^5-"a"*z^2-1=0\n");
print("Galois group: "polgalois('z^5-a*'z^2-1)"\n");
print("Ordinary solution:\n"polroots('z^5-a*'z^2-1)"\n");
print("Not-ordinary solution:");
Z= [];
for(j=0, 4,
z= exp(2*j*Pi*I/m) + 1/(2*Pi*I)
*(exp((2*j+1)*Pi*I/m)*intnum(t=0, oo, log(1+a*t^n/(1+t^m)*exp((2*j+1)*Pi*I*n/m)))
- exp((2*j-1)*Pi*I/m)*intnum(t=0, oo, log(1+a*t^n/(1+t^m)*exp((2*j-1)*Pi*I*n/m))));
Z= concat(Z, [z])
);
print(Z)
Output:
Quintic: z^5-7*z^2-1=0
Galois group: [120, -1, 1, "S5"]
Ordinary solution:
[1.9369100453804415363610723955778268241 + 0.E-38*I,
0.0014571193250340581699295533533713515578 - 0.37793919029580108279238966671308590927*I,
0.0014571193250340581699295533533713515578 + 0.37793919029580108279238966671308590927*I,
-0.96991214201525482635046575114228476362 - 1.6351464511815856113226711169592030156*I,
-0.96991214201525482635046575114228476362 + 1.6351464511815856113226711169592030156*I]~
Not-ordinary solution:
[1.9369100453804415363610723955778268026 + 0.E-39*I,
0.0014571193250034346617831783697118620865 + 0.37793919029582726259430943508563428674*I,
-0.96991214201522420284231937615862526337 + 1.6351464511815594315207513485866546421*I,
-0.96991214201522420284231937615862526336 - 1.6351464511815594315207513485866546421*I,
0.0014571193250034346617831783697118620703 - 0.37793919029582726259430943508563428672*I]
There is intutive explanation given here through Galois Theory
you have to take Galois group of the equation, the group of automorphisms of the splitting field of the polynomial. For polynomial of any degree $n$. By defination of Automorphism, it maps root of a equation to the other root, it is a subgroup $G$ of the $S_n$ (symmetric group).
Galois Theory assigns, to each polynomial, a mathematical structure called a group. A polynomial is solvable in radicals (that is, you can write down its roots in terms of its coefficients, the 4 arithmetical operations, and square roots, cube roots, etc.) if and only if the corresponding group is a "solvable" group. The definition of solvable group won't mean much to you if you haven't done a course in group theory; there should be a sequence of groups, starting with the trivial group and ending with the group corresponding to the polynomial, such that each group in the sequence is a "normal" subgroup of the next group, and the "quotient" of each group by the previous group is "commutative". But for $n\ge 5$, and of course, not abelian. Hence $S_n$, which is the Galois group is not solvable.
