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We can solve (get some kind of answer) equations like:

$$ ax^2 + bx + c=0$$

$$ax^3 + bx^2 + cx + d=0$$

$$ax^4 + bx^3 + cx^2 + dx + e=0$$

But why is there no formula for an equation like $$ax^5 + bx^4 + cx^3 + dx^2 + ex + f=0$$

I'm not sure if this has anything to do with the Galois theory, but is there a dumbed-down simple explanation as to why degree 5 polynomials (and up) are unsolvable?

Martin Sleziak
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Pacerier
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    ... unsolvable by radicals (and arithmetic), you mean. The short answer is something to the effect of studying the ways in which you can construct numbers as roots of polynomials, and how such constructions can be built from elementary, irreducible constructions. Then, you observe one of those irreducible constructions is not of the form "take the $k$-th root of some number". (where $k$ is a positive integer) –  Jul 29 '12 at 19:36
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    Also see this answer: http://math.stackexchange.com/a/792/221 – Matt E Jul 30 '12 at 02:10
  • trb456 already gave you an answer on why one can't use radicals for representing general solutions to polynomials of high degree. I'll just add the note that if you allow nonelementary functions like theta functions or hypergeometric functions, one can certainly represent the roots of those polynomials. This is not unlike the casus irreducibilis for the cubic equation, where trigonometric or hyperbolic functions are required if one wants to avoid complex numbers in representing roots that are supposed to be real. – J. M. ain't a mathematician Jul 30 '12 at 03:15
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    Because $A_5$ is simple. I skipped some details... – D Wiggles Dec 19 '13 at 01:37
  • I think I would put it this way: The kinds of expressions you can build by starting with rational numbers and nesting radicals and arithmetic operations always have a certain "symmetry" that is determined by the step-wise process you use to build them (i.e., taking successive nth roots of expressions you already obtained). On the other hand, the roots of polynomials of degree 5 or higher can exhibit a type symmetry that can not be "reached" with those building blocks. – Nicolas Miari Aug 15 '16 at 06:43
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    Nice question, pal ! – Spectre Nov 08 '20 at 08:46
  • @Pacerier, see this : https://math.stackexchange.com/questions/3829335/a-general-way-of-finding-the-zeroes-of-a-polynomial – Spectre Nov 08 '20 at 08:51

5 Answers5

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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:

AAB + BBC + CCA

ABB + BCC + CAA

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:

AB + CD

AC + BD

AD + BC

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.

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 more group theory for that. But as far as the "simple" explanation of why you can't solve the fifth degree, it's really all about those functions.

To elaborate on this in a little more detail: 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 simple 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.

This was fairly well understood by Lagrange fifty years before Galois theory made it "rigorous". It has to do with 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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    +1 for very simple explanation. If you proof why we cannot find four functions in A, B, C, D, and E which is preserved under permutations of those five letters, That will be great answer for simple explanation. I think Lagrange shew that in his book. Please check in Jim Brown's paper on page 5 .http://www.math.caltech.edu/~jimlb/abel.pdf . It is very interesting subject. I also have a question about the permuations. Maybe you can wish to check it.http://math.stackexchange.com/questions/120692/number-of-distinct-fx-1-x-2-x-3-ldots-x-n-under-permutation-of-the-input – Mathlover Dec 19 '13 at 08:14
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    I'm glad you liked it. Yes, I looked at the Jim Brown article and it is very good but I couldn't get past the 2nd line of his proof. Just not smart enough. I especially liked the part at the end where he admits "I don't know what Abel is doing here." You don't see that very often. I also checked out your question...yes, its a good question and I'm going to try to put something in on it. You might want to read some of my blogposts at http://marty-green.blogspot.ca/2012/06/fifth-degree-part-1.html – Marty Green Dec 19 '13 at 18:37
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    Thank you for this explanation. Is it difficult to explain how the 2 formulas for p and q for the 3rd degree equation (and the 3 formulas for p, q and r for the 4th degree equation) are obtained? – zar Jan 01 '16 at 15:38
  • You should check out some of my blogposts which I linked above. But to your specific question about the formulas for the p's and q's, do you understand how any function symmetric in the roots can be easily found from the elementary symmetric functions which are the coefficcients of the polynomial? So for a quadratic equation where the roots are a and b, you can easily look at the equation and tell what is a^2 + b^2? – Marty Green Jan 01 '16 at 16:39
  • I know that $a^2+b^2 = (a+b)^2 - 2ab$ – zar Jan 01 '16 at 17:56
  • Oh, I understand your question now: by looking at $Ax^2+Bx+C=0$ I can tell that $-B/A=a+b$ and $C/A=ab$, so $a^2+b^2=(-B/A)^2-2C/A$. – zar Jan 01 '16 at 19:11
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    Yes, that's right. By the way, for me it makes things a little simpler if I let all my polynomials be monic (leading coefficient A=1) so I can just think: the product of the roots is the constant coefficient, and the sum of the roots is the (negative) coefficient in x. Anyhow, you see how the coefficients of the polynomial are the ELEMENTARY symmetric functions of the roots, and the derived functions (my p's and q's) are (general) symmetric functions. Which can always be worked out from the elementary functions. – Marty Green Jan 02 '16 at 03:06
  • Ok, thank you. I can't figure out if this procedure has something to do with the commutators used in the Arnold's proof video cited before. – zar Jan 02 '16 at 08:15
  • No I don't think it does. I'm in much more traditional territory. The Arnold proof comes from a totally novel perspective. – Marty Green Jan 02 '16 at 15:33
  • I thought there was a connection, because you mentioned a 60 terms formula, and the Arnold proof shows that the cardinality of the set of commutators is 60 – zar Jan 02 '16 at 19:25
  • "If you try to create functions on five letters with this symmetry property, you'll convince yourself that it's impossible. " I think I read somewhere that there are analogous symmetric for the quintic, but they yield a polynomial of the 6th degree (therefore, the "stepping stone" lies higher than the goal and is of no help) – Nicolas Miari Aug 15 '16 at 06:38
  • Yes, you're right in a way. There is a set of functions called Dummit's Resolvents which we discussed this in another thread which I think you can find here: http://math.stackexchange.com/questions/615125/resolvent-of-the-quintic-functions-of-the-roots – Marty Green Aug 15 '16 at 06:50
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I'll try a "dumbed down" version, although @Robert Israel's answer plus comments are fine!

Solvable means solvable by radicals, and that means that, starting from the polynomial equation, you can only do 1) field arithmetic $(+,-,\times,\div)$, or 2) "extracting roots; e.g. square roots, cube roots, etc. It is the case, by Abel-Ruffini first and then by Galois, that there is no general "formula" for solving polynomials above degree 4. Naively, that suggests that the formula gets "too complicated" at some point. @paul garrett gets at this when he refers to the resolvent, which is a step that can simplify solving if the resolvent polynomial is of lower degree.

Galois found that the way to measure "too complicated" is by checking which roots of the polynomial can be "switched around", or permuted, while maintaining certain equations of the roots. For example, if you are working over the rational numbers, then you can't switch around any rational number without changing important relationships. That seems obvious. But what might seem strange is that for a polynomial like $x^2-2$, whose roots are $\sqrt2$ and $-\sqrt2$ , you can switch these around and not hurt any other arithmetic!

The way to formalize what it means to "switch around" roots is thought group theory, and there is a group that corresponds to how the roots of a polynomial can be switched around called the Galois group. Finally, if this group is "too complicated" (i.e. too many ways to permute the roots), then that group and its corresponding polynomial are not solvable by radicals. In the case of 5th degree polynomials, if it were possible to "invert" the polynomial $x^5-x-1$ (i.e. solve it directly like we can $x^5$), I believe this is all that would we needed for all 5th degree polynomials to be solvable by radicals. So as you see, it's just a "little bit" too complicated, and it gets worse as the degree increases.

I'm leaving out lots of details, but the other answers and links fill in those details. But I hope this gives you a flavor if what's going on.

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    Can't be $x^5-x$ since that has trivially $0$, $1$, $i$, $-1$ and $-i$ as solutions. – Bruno Le Floch Jul 29 '12 at 22:48
  • Thank you--meant $x^5-x-1$, fixed. –  Jul 29 '12 at 23:06
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    I disagree with the need to fix. To invert the polynomial $x^5-x$ already means to solve the equation $x^5-x=c$. I believe that's exactly what you meant already. I don't think the claim is true if "invert" is interpreted in the limited sense that @BrunoLeFloch implies. – Erick Wong Jul 29 '12 at 23:31
  • @Erick Wong--Yes, but I think the edit makes this clearer. What you say is still true after the edit. –  Jul 29 '12 at 23:36
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    @ErickWong My bad. You are right, let me retract my comment. Perhaps 'if it were possible to "invert" the polynomial $x^5−x$ (i.e. solve $x^5-x=c$ directly like we can $x^5=c$)' would be better? – Bruno Le Floch Jul 30 '12 at 00:19
  • @Bruno Le Floch: This is exactly right, and what I intended, but the edit is equivalent and does eliminate any confusion. –  Jul 30 '12 at 00:29
  • Thus for inverting $x^5-x=c$, we require what is called the Bring(-Jerrard) radical, which is not a radical in the usual sense... – J. M. ain't a mathematician Jul 30 '12 at 03:18
  • @trb456, do you mean that they are solvable but just that so far no formula is found yet, and in the near future we may very well be able to solve equations of the form ax^5+bx^4+cx^3+dx^2+ex+f=0 ? – Pacerier Jul 30 '12 at 03:20
  • @Pacer, he meant that the roots of a general quintic cannot be expressed in terms of radicals, and their sums/products/quotients. As I commented in your question, if you allow the use of certain special functions (the Bring-Jerrard "radical" among them; look it up!), then one can have "closed form" expressions for the roots. – J. M. ain't a mathematician Jul 30 '12 at 10:54
  • @Pacerier: No, under the rules I gave (field math and extracting roots), there is NO general formula for polynomials of degree 5 and higher. $x^5-x-1$ is an example. Galois theory proves this rigorously; i.e. a polynomial is solvable iff its Galois group is solvable, which captures this notion of "complexity". –  Jul 30 '12 at 10:54
  • I should add: you can approximate the roots of any polynomial. Go to Wolfram Alpha right now and have it solve $x^5-x-1$--it will! But it's using numerical methods to do so. Solvability by radicals is more about the "deep structure" of polynomial rings. –  Jul 30 '12 at 11:00
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It has everything to do with Galois theory, although the original proof preceded Galois.

See this Wikipedia article on Ruffini theorem.

I don't believe there is a "dumbed-down simple explanation".

Gigili
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Robert Israel
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    +1 for not attempting a dumbed-down version of the historical arguments. – Old John Jul 29 '12 at 19:36
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    ... and for quintics, unlike lower-degree, the relevant Lagrange resolvent is of higher degree, rather than lower, so (generically) that approach fails. Of course, failure of one approach does not prove impossibility, which is what Ruffini, Abel, and Galois proved. – paul garrett Jul 29 '12 at 19:39
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The topological proof by Arnold is simpler to understand. A shortened version is explained in http://www.youtube.com/watch?v=zeRXVL6qPk4 This requires (only) knowledge of complex numbers and their roots.

Boaz Katz
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  • Not a bad explanation perhaps. Sort of holistic, but it is a start I feel. – Vladhagen Dec 07 '13 at 03:34
  • Although we would have to watch the half of an hour video to begin to grasp the concept. – Vladhagen Dec 07 '13 at 03:40
  • Not sure what do you mean by "holistic", but I consider this the perfect answer. In particular, it can be completely formalized (unless I am missing something, and if you ignore singularities in division which I believe can be handled by compactification). – Zuza Jun 01 '17 at 21:22
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There is already one "dumbed" answer above, so forgive me for adding an even "dumber".

What it means by solvable is to be able to write down a formula for a solution of a polynomial equation like $ax^2 + bx + c = 0$ such this one:

${x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}$

In other words, we need the solution (whatever it would be in $x$) could be expressed in some algebraic expressions. In the example above: $-, b, \pm, \sqrt, \: ^2, 4, a, c, /,$ and $2$ are all algebraic expressions.

For most of all numbers we know, they could be expressed in algebraic expressions. For example: $2, 6/3, \sqrt{2}, \sqrt{-1},$ etc and even for a non-repeating irrational number like $\pi$ could be expressed in $\frac{C}{D}$ where $C$ is the circumference of a circle and $D$ is the diameter.

Amazingly there are numbers that could not be expressed in any algebraic expressions we know.

Historically, it was discovered when observing that polynomial equations of degree higher than 4 may not necessarily have a solution that could be expressed in algebraic expressions. It was first observed by Joseph-Louis Lagrange in 1770, partly proven by Paolo Ruffini in 1799, and then completed by Niels_Henrik_Abel in 1824, establishing Abel–Ruffini theorem. But not until 1830 Évariste Galois at the very young age of 18 showed the necessary condition of a polynomial equation to have a solution hence called solvable.

The theory is called Galois theory, but unfortunately it is too complex to be explained satisfactorily here. If we have enough background in fundamental algebra, it would still need at least 200 hundreds pages of hard math to understand it. Nevertheless the necessary condition showed by Galois theory also shows that there are polynomial equations that their solutions could not be expressed in any algebraic expressions.

An example of a such unsolvable polynomial equation is $x^5 - x - 1 = 0$. Note to not be confused with the algebraic expression mentioned above. In the equation $x^5 - x - 1 = 0$, we cannot transform it into a solution expression where a single $x$ is put on the left side of the equation such $x = \: ...$ because, well, we don't have the algebraic expression to be written down in the right side of the equation.

Of course we could approximate the solution with some number close enough, but it is still NOT the solution to the equation because simply it is just close enough BUT not the exact solution. The exact solution is not expressible yet. Amazing right!

Ari Royce Hidayat
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