There is no solution in radicals to general polynomial equations of degree five.

Question

I have read that there is no formula for the quintic equation.

Tow questions:

1- What does that even mean? By the fundamental theorem of algebra, any equation of degree n has n solutions in the complex numbers, so any quintic equation has 5 solutions that either can be written in radical form or not. If not, how the solutions belong to the complex numbers? are there complex numbers that can't be written in radical form?

2- How it's even possible to prove such a statement? It seems that you must try every possible combination of symbols and mathematical operations, and for each combination construct a formula, and for every constructed formula -there is an infinite number of possible formulas-, you have to prove it's false.

Can you give me an example of a quintic equation with no radical solutions?

Also, the equation $x-pi=0$ has no radical solutions, but that doesn't imply that there is no general formula for the first-degree equations!

Answer 1

First of all, there are two subtly different versions of the question you're asking, and you may not be aware of the difference:

1. Are there quintic polynomials which have roots which cannot be expressed using radicals?
2. Is there a general formula which simultaneously solves all quintic polynomials?

In point (1), we permit using a different formula in radicals to solve each individual quintic, or even a different formula to express each root of each quintic. We can get to the heart of the matter by dropping polynomials from the picture: it is a theorem that there exist real numbers $r$ which cannot be expressed as formulae in radicals using rational numbers. We can prove this on set-theoretic grounds: there are only countably many formulae in radicals using rational numbers.

Below I'll address version (2) of the question, although we'll deal with version (1) on the way. I'm going to define, recursively, a certain set of "formulas". If you find it more concrete, you can identify a "formula" with a tree, the nodes of which are operations and the leaves of which are the symbols $a_1, a_2, a_3, a_4, a_5$ and any rational number.

1. Each rational number is a formula.
2. Each symbol $a_1, a_2, a_3, a_4$ and $a_5$ is a formula.
3. If $f$ and $g$ are formulae, then $f\cdot g$ is a formula.
4. If $f$ and $g$ are formulae, then $f+g$ is a formula.
5. If $f$ is a formula, then $\sqrt[n]f$ is a formula, for any positive integer $n$.

We can then define $F_5$ to be the intersection of all sets of formulae satisfying the above. This is the set of "solutions in radicals" for a polynomial of degree $5$. Of course the $a_i$ are to be interpreted as the coefficients of the polynomial.

Given such a formula, we can "evaluate" it for specific value of the $a_i$. But we have to be careful: notice that in general there isn't a unique value for a formula when we plug in values for the $a_i$, because $n$-th roots are multi-valued. When we nest $n$-th roots, we get a combinatorial explosion of different possible values. We can recursively define what it means for a complex number $z$ to be "a value" of a formula for given values of the $a_i$ - but not "the value".

I can now state my answer to question (1) more rigorously. If we drop axiom (2) from the above list, we get a certain more restricted set of formulae $F_5'$, formulae with no variables $a_i$ in them - only concrete rational numbers. As discussed above, we can define what it means for a complex number to be "a value" of a formula in $F_5'$. $F_5'$ is countable and each formula only has finitely many values. Therefore there must exist real numbers $r$ such that $r$ is not a value of any formula in $F_5'$. It is furthermore the case that some of these real numbers are roots to quintic polynomials in rational coefficients, but this is harder to prove.

For question (2), to say a given formula $f$ in $F_5$ is a general solution to the quintic means that for any assignment of values to the $a_i$, each root of $a_1+a_2x+a_3x^2+a_4x^3+a_5x^4+x^5$ is a value of $f$ when we plug in those values. There is no such formula $f$.

The answer to the question how do we prove this is basically "Galois theory". I wish I understood the proof deeply enough to be able to give you the general idea, but I don't. In any case, once we have a definition for what it means for a formula to solve the quintic, it shouldn't be surprising that something can be proved about that definition.

Answer 2

Are there complex numbers that can't be written in radical form?
Of course. In fact there are even real numbers that cannot be written in radical form (in terms of rational numbers). Such as $\pi$ and $e$.

There is no solution in radicals to general polynomial equations of degree five.
A theorem of Abel, improved by Galois. Mathematics students learn about this under the heading Galois Theory. See Wikipedia

added
Of course $x-\pi=0$ has solution $\pi$. Indeed, any complex number $a$ is a solution of $x-a=0$. But I intend to say: no polynomial equation with rational coefficients has $\pi$ as a solution.
A more complete statement:
Let $$ x^5 + A x^4 + B x^3 + C x^2 + D x + E = 0 \tag1$$ be your quintic equation. There is no finite formula for a solution of $(1)$ in terms of the coefficients $A,B,C,D,E$ using the operations: addition, subtraction, multiplication, division, radicals.

Answer 3

Can you give me an example of a quintic equation with no radical solutions?

Try the following example. Find a formula by radicals for the solutions of $$ X^5-4X+2=0. $$ The Galois group of this polynomial over $\Bbb Q$ is ismorphic to $S_5$. This group is not solvable. Hence by Galois the polynomial equation is not solvable by radicals.