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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?

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You didn't write "$=0$" after these polynomials, so what you've got are not actually equations. – Michael Hardy Jul 29 '12 at 19:36
... 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) – Hurkyl Jul 29 '12 at 19:36
Also see this answer: – Matt E Jul 30 '12 at 2:10
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. Jul 30 '12 at 3:15
up vote 11 down vote accepted

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. – trb456 Jul 29 '12 at 23:06
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. – trb456 Jul 29 '12 at 23:36
@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 0:19

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".

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+1 for not attempting a dumbed-down version of the historical arguments. – Old John Jul 29 '12 at 19:36
... 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

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.

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.

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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 . It is very interesting subject. I also have a question about the permuations. Maybe you can wish to check… – Mathlover Dec 19 '13 at 8:14
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 – Marty Green Dec 19 '13 at 18:37
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 1 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 1 at 16:39
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 2 at 3:06

The topological proof by Arnold is simpler to understand. A shortened version is explained in This requires (only) knowledge of complex numbers and their roots.

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Not a bad explanation perhaps. Sort of holistic, but it is a start I feel. – Vladhagen Dec 7 '13 at 3:34
Although we would have to watch the half of an hour video to begin to grasp the concept. – Vladhagen Dec 7 '13 at 3:40

Because $A_5$ is simple. I skipped some details...

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