# Intuitive reasoning why are quintics unsolvable

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.

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:

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


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.