I understand that the quadratic equation can solve any second order polynomial. Furthermore, equations exist for polynomials up to fourth order. However, without a graduate level degree and a deep understanding of mathematics, is there an explanation of how we can prove that the equations for solutions to a fifth order and above don't exist?

Thanks in advance.

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    $\begingroup$ Radical formulas for fifth order don't exist. $\endgroup$ Commented Nov 8, 2014 at 0:41
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    $\begingroup$ youtu.be/cxNq-hQwvn0 does a nice version of the explanation (1 hr). And youtu.be/RhpVSV6iCko provides a visual. $\endgroup$ Commented Nov 9, 2014 at 5:32
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    $\begingroup$ One should note that the galois result involves using radicals. For a little more intuitive explanation (if i may say) take a look here, plus for other types of solutions of quintics take a look here $\endgroup$
    – Nikos M.
    Commented Nov 12, 2014 at 12:17

4 Answers 4


Here's a very vague intuition for why such a proof might be possible.

Many people are shocked when they hear that the general quintic cannot be solved. "What does that mean?", they say. "Just because many people have tried and failed does not mean that's impossible. Flight was impossible until the Wright Brothers took off at Kitty Hawk...", etc. etc. But when we talk about solving the quintic, we are not talking about as some Herculean task that only a genius can accomplish. We're just talking about something that, as it turns out, is not allowed according to the rules of the algebra game.

One way of proving that a given task is impossible is to find something that stays the same when you perform any of the possible "moves" that are allowed. For example, suppose you were teaching someone the rules of checkers, you play a couple games, and then you point out that a piece on a black square can never find its way to a white square. "What?!", your friend bellows. "I don't see why not! Flight was impossible until...", etc. But all you need to do to prove this simple fact is to observe that at any given time, none of the rules allow you to move from black to white.

In the algebra game, you're given a polynomial $p(x) = a + bx + cx^2 + dx^3 + ex^4+fx^5$ and you're allowed to manipulate the symbols $a,b,c,d,e,f$ and numbers $1,2, \ldots$ in various ways. For example, you can add $a$ and $b$ to get $a+b$; you're allowed to multiply, divide, etc. This gives you what mathematicians call a field.

You're also allowed to use the symbol $\sqrt{}$, so to take square roots, cube roots, etc.. This lets you make complicated-looking expressions like $$\sqrt[8]{\frac{a + b \sqrt{d +e^3}}{e-\sqrt[4]{5}}}.$$ But to do this, you have to make your field bigger. This is called a field extension. As it turns out, there's a certain property that's always the same when you make field bigger by taking roots. It's a little complicated to explain what - for that you must learn Galois theory. But it's a little bit like the way a checker can't change colors.

It turns out that for certain polynomials $p(x)$, you can prove that any field that contains any root of $p(x)$ doesn't have that property. That means that you can spend all day playing with roots, etc., and you will never solve the general quintic. A clever mathematician like Évariste Galois can see the whole game at once and realizes that your task is not possible.

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    $\begingroup$ This is really well written. $\endgroup$
    – jtbandes
    Commented Nov 8, 2014 at 6:35
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    $\begingroup$ So it takes more of a broad perspective to see? $\endgroup$ Commented Nov 8, 2014 at 18:00
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    $\begingroup$ Wait, so if we define some other functions that expand the playing field (for example, fifth root is inverse to $x^5$, so inverse to $x^5 + x$?) then it may be possible to solve quintic polynomials? Is there a certain function that we can define that allows us to find a formula for all polynomials? $\endgroup$
    – soktinpk
    Commented Nov 9, 2014 at 2:21
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    $\begingroup$ @soktinpk: that depends on how wild you want to go. Polynomials exist, their roots (when they do) exist, so of course there is a function. The problem is that you want a function that is a formula on the coefficients of the polynomial. And beyond doing the basic operations and taking $n^{\rm th}$ roots, what else there is? $\endgroup$ Commented Nov 9, 2014 at 4:36
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    $\begingroup$ @soktinpk: The following might be of interest regarding quintics: math.stackexchange.com/questions/540964/…, en.wikipedia.org/wiki/Quintic_function#Beyond_radicals $\endgroup$ Commented Nov 12, 2014 at 0:36

A common proof is as follows:

It can be shown that, over a field of characteristic 0 (like the rationals, the reals, or the complex numbers) there exists a formula for the roots of a polynomial $p(x)$ in terms of radicals (square root, cube root, etc.) if and only if the Galois group of its splitting field (the field extension at which $p(x)$ can be expressed as a product of linear factors) over the base field is solvable. The solvability condition intuitively means that the group can be successively broken down into a series of smaller groups in a systematic way eventually leading to a group with one element, and each term in this series corresponds to taking the $n$th root of some element in a sequence of successsively larger fields leading up to the splitting field.

If there were a general radical formula for solving equations of fifth degree, then for arbitrary algebraically independent transcendental elements (the opposite of algebraic elements, meaning they are not the solutions to any polynomial equation over the base field) $a$, $b$, $c$, $d$, $e$, and $f$, the splitting field of the polynomial $ax^5+bx^4+cx^3+dx^2+ex+f$ should have a solvable Galois group.

However, the Galois group of this polynomial is $S_5$, the symmetric group on five letters, which is not a solvable group. Therefore there exist no radical formulas for general fifth degree equations, and hence no such formula for equations of higher degree either.

However, this does not mean that it is not possible to solve any polynomial equations of higher degree. There exist polynomials of higher degree that have solvable Galois groups, so their roots can be expressed with radicals. There is just no general formula that works for all equations of a given degree greater than or equal to 5. There are also specific polynomials with integer coefficients whose splitting fields do not have solvable Galois groups.

  • $\begingroup$ Do you have reference where this statement of yours is explained maybe with examples, "There exist polynomials of higher degree that have solvable Galois groups, so their roots can be expressed with radicals" ? $\endgroup$ Commented Mar 9, 2018 at 20:39
  • $\begingroup$ @gradstudent what about $x^5-1$? It's easy to construct examples. $\endgroup$ Commented Mar 16, 2018 at 0:30
  • $\begingroup$ Sure! I meant a proof of this general statement that you say. Some from-scratch exposition of this proof would be helpful to me! $\endgroup$ Commented Mar 16, 2018 at 0:57

I think you're asking someone to argue why this is plausible, not why it's true.* So I won't prove the result; that can be found on the Web as well as in books. Instead I'll try to make it sound "possible in theory" to do what was done. And I won't use any words above GCSE level.

First some background. This general statement ("not even logically possible to do X") came from a new style of abstract and also creative thinking. It's been said that modern mathematics begins with Galois, if not in the sense of increased rigour then at least Galois demonstrated the promise of general thinking and invention of new concepts. His invention of abstract group theory did not prove a new fact; the Abel-Ruffini theorem was decades old when he wrote his memoir. Instead, abstract group theory made the entire subject of $N$th order polynomials, make a lot more sense to anyone who understood him.

Also, Galois theory combines numbers, algebra (symbol manipulation), and, in some sense, geometry (mental pictures). Cross-topical work is valued in modern mathematics ever since.

galois tower of fields

So how can it be possible to prove that not-all-quintics can be solved with one universal $+, -, \times, \div, \sqrt{}$ formula?

  • Playing around with polynomials in the 1700–1800's, people noticed a few things. First they noticed that radicals reduce symmetry. $\sqrt{\bullet}$ reduces their symmetry by $\div 2$; $\sqrt[3]{\bullet}$ reduces their symmetry by $\div 3$, and so on.

  • People also looked at when polynomial equations are solvable-in-general. In particular they looked at how symmetry-reduction affects solvability. They proved that in order for a polynomial with coefficients $a,b,c,d,e,f,g$ to be solvable-by-radicals, the symmetry of the letters $a,b,c,d,e,f,g$ must act a certain way.**

They then proved that the letters $a,b,c,d,e$ do not act in the required way. If

  1. their symmetry is not like this, and
  2. their symmetry would need to be like this in order for $ax^5+bx^4+cx^3+dx^2+e=0$ to be solvable-by-radicals, then
  3. $ax^5+bx^4+cx^3+dx^2+e=0$ can't be solvable-by-radicals.


So to review: people looked at symmetry properties of things, and solvability properties of things. They related solvability to symmetry. Then they proved something about symmetry. That gave them a fact about solvability.


* A truth being implausible just makes it more interesting! I like this quotation (which I believe is due to Andrei Kolmogorov): "Mathematics occurs on the boundary between the obvious and the impossible".

** The particular symmetry they must display, and the relationship between solutions and symmetry, and the theory of symmetries in general, is what requires the education you said you don't want to need to understand this answer.

  • $\begingroup$ Does it means that, if we use more tools, not only +,−,×,÷,√, we could solve the polynomial more than 5 degrees by radical? $\endgroup$
    – Thaina
    Commented May 21, 2018 at 11:57
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    $\begingroup$ @Thaina It depends on the tools you are using. If you use only radical tools (square-root and n-th-root), you could not. If you use other kind of tools (elliptic functions), yes you could solve polynomial more than 5 degree. But this would not be solving by radical anymore... $\endgroup$
    – Florian
    Commented Sep 25, 2020 at 11:49

At one time I had the idea that if there wasn't one formula perhaps there was a variety of formula that could be selected from; even infinity one of an infinite.

The class I took explained it very carefully.
First it started with a straight edge and compass and computed the domain of calculations that could be done with them; and then showed trisecting the angle was not in that domain. Then the book went on to construct the domain of radical solutions and showed it didn't include certain roots of fifth order polynomials.

Now there are certainly solutions; but the point is they are not even in the domain of radical algorithms. No collection of formulas work because none of them can even reach those roots.
And as mentioned above Galois showed clearly which formulas could be solved and which couldn't.

It's like expressing pi as a rational number; you might get close but are never going to get there for a finite rational.

  • $\begingroup$ As if we were approximating a solution? Do you know how close the approximate answer was to any definite solution? $\endgroup$ Commented Nov 12, 2014 at 21:46
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    $\begingroup$ Even the rationals can approximate any number rapidly. And rationals are a subset of the fields generated with algebraic numbers. For a start on apptoximation speeds try:en.wikipedia.org/wiki/Diophantine_approximation $\endgroup$
    – rrogers
    Commented Nov 12, 2014 at 23:10

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