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.

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    I don't think it's true that any polynomial can be brought to the form $ax^n + bx + c$. Quintics can, though; see http://en.wikipedia.org/wiki/Ultraradical. – mjqxxxx Nov 03 '13 at 17:40
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    @Sawarnik : this is a good question, but I don't think there is an intuitive answer. The actual proof that quintics are unsolvable is quite difficult, and I doubt there is a simple intuitive reason why this is so. – Stefan Smith Nov 03 '13 at 17:47
  • @StefanSmith I know that, but at least but I hope for just a reasoning, a tickle in that direction, just why it 'should' be so. – Sawarnik Nov 03 '13 at 17:52
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    A nice account of Abel's original proof (as well as its history) is http://www.math.caltech.edu/~jimlb/abel.pdf‎ . But it also doesn't mention any intuition. My personal intuition would use Galois theory and that $S_n$ is solvable for $n \leq 4$ since it's "too small", but $S_5$ is not small enough anymore. But I think that this won't satisfy because it's not precise and it uses Galois theory. – Martin Brandenburg Nov 03 '13 at 17:57
  • http://podgallery.org/why-beauty-is-truth-short/. This link is to Ian Stewart's podcast. There are 7 15 minute episodes where he builds up the theory of quintics through Babylon, to Cardano, and then Galois. It is not "mathy" but I think very informative to the non undergraduate to understand why it works the way it does... – Eleven-Eleven Nov 03 '13 at 17:59
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    I should note that the piece on Galois is the 4th in the series so you don't have to wait long to hear about it. – Eleven-Eleven Nov 03 '13 at 18:08
  • The closest I've seen to an intuitive explanation was a lecture that talked about a visual understanding of why the quintic is insoluble. From the abstract: "You have probably heard that there is no formula for the roots of a degree five polynomial. I'll explain what this means and how we can prove it be looking at how the roots of such a polynomial change as we vary the coefficients." He let a coefficient of the polynomial traverse a loop in the complex plane, and observed the movement of the roots of the polynomial as the path was traversed. – Stahl Nov 03 '13 at 18:15
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    (con't) It turns out that while the roots will be sent to roots, they do not necessarily return to their original locations! Keeping track of that data via permutations, he was able to show that the Galois group in question was unsolvable. However, this still requires basic knowledge of Galois theory and group theory. – Stahl Nov 03 '13 at 18:15
  • Actually, this isn't the same person, but [this](http://ifandifonly.wordpress.com/2011/11/17/monodromy-insolubility-of-the-quintic-and-enumerative-problems-in-geometry/) blog post discusses the same idea. Interesting stuff, but I fear too high-brow to be an answer to this particular question. – Stahl Nov 03 '13 at 19:37
  • There is an interesting paper by D. S. Dummit called " Solving Solvable Quintics" . The algebraic manipulations used in the paper are of the same order of difficulty as those used in handling the general cubic. The paper also explains why certain Quintics are solvable, ( The Galois group is contained in the Frobenius group of order 20.) My premise for posting is that when you compute something the subject becomes conceptually easier to understand and your intuition improves. I'm sure Galois would agree... – Alan Nov 03 '13 at 19:51
  • @Alan Can you give me an link? Seems to be interesting. – Sawarnik Nov 03 '13 at 20:01
  • http://www.ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/S0025-5718-1991-1079014-X.pdf – Alan Nov 03 '13 at 20:04
  • @Stahl, I believe you're thinking of David Speyer's talk at the 25th PROMYS anniversary. For intuition, I remember Brian Conrad wrote something here or on MO about how solving for a root of a quintic requires breaking some existing symmetry between the roots, and when the symmetry group of the roots is too complicated (not solvable), this can't be done in terms of radicals. I can't find the original post, but if someone can that might be helpful. – Dylan Yott Nov 04 '13 at 05:28
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    I can't help with much with the intuitive reasoning, but I can attempt to give a very brief summary of a very small part of the content of Ian Stewart's book on the subject (called 'Galois Theory'): none of the tricks which can be used to solve polynomial equations of degree 1-4 work for quintics. 'Lagrange analysed all of these tricks [in 1770-71] and showed that they can all be explained using general priciples about symmetric functions of the roots. When he applied this method to the quintic, however, he found that it reduced the problem to solving a sextic- an equation of degree 6. – George Tomlinson Nov 04 '13 at 17:46
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    (cont.) Instead of helping, the method now made the problem *worse*'. A lecture covering this material by George Neville Watson was written up by Berndt, Spearman and Williams (2002). Although Lagrange's method fails for the quintic, an impossibility proof is needed in order to be certain there are no other tricks for solving them. It turns out that in order for a given polynomial equation to be soluble by radicals, the associated Galois group must be a soluble group. Stewart then provides an example of a quintic whose Galois group isn't soluble: $t^5−6t+3=0$. Galois died in a duel, aged 22. – George Tomlinson Nov 04 '13 at 17:46
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    If there is a simple intuitive reason why quintics are not solvable, then it eluded the world's best mathematicians for decades, didn't it? I thought people spent years trying to find formulae for solutions, before Abel finally showed that the hunt was destined to fail. In fact, some determined souls are still looking for the magic formulae :-) – bubba Mar 06 '14 at 12:22
  • @bubba Yes, I see what you say. I was just asking whether there were some reasoning that the prodigious Galois must have had before working and proving his theory. But, now I think its very much hopeless to go behind it. I think I will start abstract algebra in a year, and see that myself :) – Sawarnik Mar 06 '14 at 15:21
  • In his comment on Sawarnik's question, George Tomlinson mentions a paper by Berndt, Spearman and Williams (2002). Here is a link where you can download the paper at no charge. https://www.researchgate.net/publication/225663875_Commentary_on_an_unpublished_lecture_by_G_N_Watson_on_solving_the_quintic – Deacon John Nov 10 '17 at 03:58

4 Answers4


Out of a lack of modesty, I am going to post my two cents worth as an answer rather than a comment, as others more qualified than me have done. Let the roots of an equation be A, B, C, etc. We are told that the unsolvability of the general quintic equation is related to the unsolvability of the associated Galois group, the symmetric group on five elements. I think I can tell you what this means on an intuitive level.

For three elements A, B, and C, you can create these two functions:


These functions have the interesting property that no matter how you reshuffle the letters A, B and C, you get back the same functions you started with. You might reverse them (as you would if you just swap A and B) or they might both stay put (as they would if you rotate A to B to C) but either way you get them back.

For four elements, something similar happens with these three functions:


No matter how you reshuffle A, B, C and D, you get these three functions back. They might be re-arranged, or they might all stay put, but either way you get them back.

For five elements, there exists no such group of functions. Well, not exactly...there is a pair of huge functions consisting of sixty terms each that works, similar to the ones I drew out for the cubic equation...but that's it. There are no groups of functions with three or especially four elements, which is what you would actually want.

(EDIT: There is also a set of six functions that map to each other under permutations, but these don't help you either. We had an intersting follow-up about Dummit's Resolvents in this discussion here Resolvent of the Quintic...Functions of the roots)

If you try to create functions on five letters with this symmetry property, you'll convince yourself that it's impossible. But how can you prove it's impossible? You probably need a little group theory for that, which I haven't yet written it up in a presentable form (but I think I can). I've written more about this on my blogsite in a series of articles starting here. You'll see I left it hanging in midstream about a year ago, but I think I'm going to finish it off soon.

I listened to the podcast which Christopher Ernst linked to in the comments, and I didn't think it was very good. Yes, it's all about the symmetries, but just because a guy is talking with an English accent doesn't mean he's profound. I'm not even sure that the stuff he said about re-shuffling five sets of 24 elements even makes sense. Anyhow, stuff on the level that Stewart is talking about was already understood long before Galois... Lagrange (most notably) had worked out all those symmetries fifty years earlier. There's an exceptionally good article about these things on a website by one Fiona Brunk which you can read here.

EDIT: I'm going to expand on the answer I posted the other day, because I think I really have identified the "intuitive" reason the quintic is unsolvable, as opposed to the "rigorous" reason which involves a lot more group theory. For the third degree equation, I identified these functions:

AAB + BBC + CCA = p

ABB + BCC + CAA = q

A, B and C are the roots of a cubic, but p and q are the roots of a quadratic. You can see that because if you look at pq and (p+q), the elementary symmetric polynomials in p and q, you will see they are symmetric in A, B and C. So they are easily expressible in terms of the coefficients of our original cubic equation. And that's why p and q are the stepping stone which gets us to the roots of the cubic.

Similarly, for the fourth degree, we identified these functions:

AB + CD = p

AC + BD = q

AD + BC = r

You can rewrite the previous paragraph word for word but just take everything up a degree, and it remains true. A, B, C, and D are the roots of a quartic, but p,q and r are the roots of a cubic. You can see they must be because if you look at the elementary symmetric polynomials in p, q and r, you will see they are symmetric in A, B, C and D. So they are easily expressible in terms of the coefficients of our original quartic equation. And that's why they are the stepping stone which gets us to the roots of the quartic.

And the intuititve reason why the fifth degree equation is unsolvable is that there is no analagous set of four functions in A, B, C, D, and E which is preserved under permutations of those five letters. As I mentioned earlier, I think Lagrange understood this intuitively fifty years before Galois. You probably needed a little more group theory to make it completely rigorous, but that's another question.

I think Lagrange would have understood the algebraic tricks whereby you went from, say, A B and C to p and q. It involves taking linear functions which mix A B and C with the cube roots of unity and examining the cube of those functions. Its a reversible process, so you can work backward the other way (by taking cube roots of functions in p and q) to solve the cubic. A very similar trick works for the fourth degree. I think Lagrange was able to show conclusively that the same trick does not work for the fifth degree...that's the "intuitive" proof. The "rigorous" proof would have had to show that in the absence of the obvious tricks (analogous to the 3rd and 4th degree), there was no other possible tricks that you could come up with.

Marty Green
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    The idea here is nice but I don't see an intuitive reason why we must get such functions, and how to get $A,B,C$ from $p,q$ for the cubic, and similarly for the quartic. – user21820 Mar 04 '17 at 03:40
  • There's no intuitive reason why those functions have to exist; it's just a fact that they exist for the cubic and quartic, but not for the quintic. If there's anything to be grasped "intuitively", it's that there between the fully symmetric functions of the roots (which are the coefficients of the polynomials) and the independent roots themselves, there exist functions of intermediate symmetry which serve as stepping stones to the isolations of the roots. – Marty Green Mar 04 '17 at 04:20
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    I understand that part of your post. What I don't understand is why we need such kind of functions of the coefficients in the first place, and how to use them to get the roots. – user21820 Mar 04 '17 at 04:24
  • It might help you to look at the form of the p's and q's for the cubic equation. They are the cube roots of numbers that look like a+/-sqrt(b). So a + sqrt(b) = p^3, and a - sqrt(b) = q^3. There are 3 values of p and 3 values of q. It looks like there should be nine possible values for p+q, but the symmetries allow you to reduce this to three. – Marty Green Mar 04 '17 at 04:32
  • Umm... I'm already familiar with the Galois theory behind the unsolvability. What I cannot see is why we **need** to find functions that are preserved under symmetry of roots, and furthermore how to retrieve the roots from such functions. – user21820 Mar 04 '17 at 04:35
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    I'm just saying that it's kind of a miracle, which people seldom bring up, that you can even solve the cubic and quartic. Because once you have a formula involving the cube root of this and the square root of that, there are already six possible values for the expression...so how do you get that down to three roots? The symmetries are critical. But that's a whole different topic that my answer doesn't directly address. – Marty Green Mar 04 '17 at 10:24
  • Just answer my question directly: For the cubic, how do you get $p,q$ from $A,B,C$ as defined in **your** post. I'm not interested in the usual solutions of the cubic because I already know those, namely Cardano's and Lagrange's. – user21820 Mar 04 '17 at 10:35
  • Okay I see. It's easy. If you add p and q, you'll see the result is a symmetric polynomial in A, B, and C. Which means it is a number which you can easily work out from the coefficients of your cubic. If you take the product of p and q...same thing. It's just a number. So you have two numbers, p and q...you know their sum, and you know their product. That's a quadratic equation and you can solve it. – Marty Green Mar 04 '17 at 10:52
  • Sorry after the back and forth you made me confused... I meant to ask (as in all my previous comments) how you obtain the **roots of the cubic** from $p,q$. I already know how to get $p,q$ from the coefficients. What your post does not explain is how to get $A,B,C$ from $p,q$. – user21820 Mar 04 '17 at 11:13
  • Erm are you able to answer my question or not: **How do you get $A,B,C$ from $p,q$?** – user21820 Mar 04 '17 at 17:49
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    Apparently I'm not able to answer your question. Sorry. – Marty Green Mar 04 '17 at 21:34
  • Okay thanks for the response. It's a pity, because if you could it would be a different solution to the cubic, and would make your answer complete! Hence I've posted a new question [here](http://math.stackexchange.com/q/2172492/21820) asking for help! – user21820 Mar 05 '17 at 04:56

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.

Paramanand Singh
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  • Dear @Parmanand, I came to this by your suggestion in one of my question. I want to understand first para of bullet; I failed to understand terms "building blocks of this complicated expression" and "intermediate expressions". For roots of quadratic equations, what these (your) terms represent in formula of root? – Maths Rahul Sep 09 '21 at 13:30
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    @MathsRahul: for the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ the radical $\sqrt{b^2-4ac}$ is a building block. If $x_1,x_2$ are roots of the equation then this radical is a rational function of the roots. You can check that the values of $\sqrt{b^2-4ac}$ are given by $\pm(x_1-x_2)$. – Paramanand Singh Sep 09 '21 at 15:21

The simplest and most intuitive reason I saw so far was:

because, when you have 5 or more things ( roots, letters, fruits ), then the ways you can permute those things can no longer be "factored" into smaller permutations that will produce the same rearrangement even when you combine that permutation with one of the other ways of permuting those 5 or more things. So for 5 and above things, no nice factorizability.

In other words, if you have four things or less, you can always "factor" those permutations into two parts:

  1. those permutations that when you combine them with any permutation from part 2 you get the same permutation regardless of whether you did the first permutation second or the second permutation first, and
  2. those permutations that are changed when they are combined with other permutations, depending on the order you combine them

So, having this nice, "factorizable" structure of the permutations makes it possible "factorize" the polynomial and to write down a nice formula in simple operations. When the number of things gets to 5 or any higher number, then the ways you can permute them don't have this nice property any longer. So you can't find any easy way to write them down with simple operations.

Stop reading here if you don't want anything more 'mathy' than this.

If you want the more "mathy" way of saying this, it is that for the group of permutations of 5 or more things, there are no normal subgroups (groups of permutations that produce the same ordering regardless of if you did that permutation first, or the combining permutation first.). When there are no normal subgroups, then the math people call this situation "simple", as in "now that is a simple group" ( when it's actually more complex, so the math people are obviously way too complex ).

If you want the "even more mathy" way of saying it, it is that $A_5$ is the smallest simple group, but that $A_1$, $A_2$, $A_3$, and $A_4$ all have nice normal subgroups you can factor out, which lets you find a nice equation to solve x in a linear, quadratic, cubic or quartic equation.

If you want a basically incomprehensible reference please enjoy visiting this Wiki entry on $A_5$.

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  • "because, when you have 5 or more things ( roots, letters, fruits ), then the ways you can permute those things can no longer be "factored" into smaller permutations that will produce the same rearrangement even when you combine that permutation with one of the other ways of permuting those 5 or more things. So for 5 and above things, no nice factorizability." Where did you see that? I'd like to follow up. – user3752935 Oct 16 '21 at 22:16

@Sawarnik, what are you referring to exactly when you write that "the formal explanation is very hard"? The basic idea is actually quite simple. I looked through the comments above and none seem to mention the smallest nonabelian simple group which happens to be the group $A_5$ of order $60$. This group is not contained in any $S_n$ for $n\leq 4$, which implies that all those are solvable, or equivalently any equation of order $\leq 4$ is solvable. Solvability of the group corresponds to the solvability in radicals of the polynomial. The lowest degree of a polynomial which makes it possible to have a simple Galois group is therefore $5$.

Mikhail Katz
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    " ... or equivalently ... " yeah, summarizing one century of mathematics in two words, as if nothing has happened :-). That's exactly what needs an intuitive explanation ... The usual proof using Galois theory doesn't need to be repeated again. This question asks for something else! – Martin Brandenburg Dec 01 '13 at 19:56
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    There is a lot of material there granted but it seems worth pointing out that the group-theoretic kernel of the theorem is the basic fact that $|A_5|=60$ and there is no smaller nonabelian simple group. This sheds light at least on the fact that $n=5$ is the "magic" degree. – Mikhail Katz Dec 02 '13 at 08:39
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    For example: didn't Abel prove unsolvability of quintics before there was any Galois theory? Maybe someone should summarize Abel's proof. – GEdgar Jan 29 '15 at 13:17