[Math] Are most rational quintics unsolvable

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It is well-known that, as polynomials of degree exceeding 4, there exist quintics whose roots cannot be solved for by radicals (Abel-Ruffini theorem). So we can divide the set of rational quintics into those which are solvable by radials and those which not. Obviously, both subsets are infinite in size.

What is not obvious to me is whether both have the ~~cardinality~~ density in rational quintics (see update below). In other words, we can ask the question: "Are there more unsolvable rational quintics than solvable?" My naive expectation would be 'yes', but I've not the background to properly formulate or prove this claim. So I'd like to see a proof either way; useful references/citations are welcome.

UPDATE: My use of 'cardinality' in my original question didn't reflect my intention. Rather, it was as MikeMiller has indicated in comments: If the subset of solvable rational quintics is denoted as $S\subset \mathbb{Q}^5$, is either $S$ or $\mathbb{Q}^5-S$ dense in $\mathbb{Q}^5$? More quantitatively, can the following limit be computed:

$$\lim_{N \to \infty} \frac{\text{# of solvable quintics with } |a_i| < N}{N^5}=?$$

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Just to get a data point, using Maple I took $2000$ random quintics with coefficients pseudo-random numbers from -100 to 100 (but the coefficient of $x^5$ nonzero). $1981$ of these were irreducible (of course the reducible ones are solvable). All $1981$ irreducible quintics were not solvable.

EDIT: Quintics with a rational root are solvable, and these are easily seen to be dense in $\mathbb Q^5$. Namely, take a rational approximation $r$ of a real root of the polynomial. Then $p(X) - p(r)$ has rational root $r$, and is arbitrarily close to $p(X)$.

EDIT: If I'm not mistaken, quintics with Galois group $S_5$ are dense in $\mathbb Q^5$. Consider the proof that $x^5 - x - 1$ has Galois group $S_5$. The same proof should apply to $p(X) = X^5 +\sum_{i=0}^4 \alpha_i X^i$ as long as

All of the denominators of the $\alpha_i$ are congruent to $1 \mod 6$.
The numerators of $\alpha_0$ and $\alpha_1$ are congruent to $5 \mod 6$, those of $\alpha_2, \alpha_3$ and $\alpha_4$ are congruent to $0 \mod 6$.

$5$-tuples satisfying these conditions are dense in $\mathbb Q^5$.

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[Math] Intuitive reasoning why are quintics unsolvable

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
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.

(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.

Number Theory – Are There More Cyclic Quintics?

Yes, there are infinitely many cyclic quintics as parameterized by the Emma Lehmer quintic

$$F(y)=y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y + 1 = 0$$

This also obeys

$$ y_1 y_2 + y_2 y_3 + y_3 y_4 + y_4 y_5 + y_5 y_1 - (y_1 y_3 + y_3 y_5 + y_5 y_2 + y_2 y_4 + y_4 y_1) = 0$$

Let $p=25 + 25 n + 15 n^2 + 5 n^3 + n^4$. Then the discriminant of $F(y)$ is

$$D = (7 + 10 n + 5 n^2 + n^3)^2\,p^4$$

Also, note that if $m=n+1$, then $n\,p=m^5 + 5m^3 + 5m - 11$. A root is given by $$y = a+b\sum_{k=1}^{(p-1)/5}\,{\zeta_p}^{c^k}$$

with root of unity $\zeta_p = e^{2\pi i/p},\,$ for some integer $a,b,c$. See this MO post for the formulas for $a,b,c$.

P.S. While $p=151$ does not belong to this family, I find that,

$$x^5 + x^4 - 60x^3 - 12x^2 + 784x + 128 = 0$$

with discriminant $d=2^{18}151^4$ has the root $\displaystyle x=\sum_{k=1}^{30}e^{2\pi\, i\, c^k/151}$ for $c=23$. The Mathematica command to find these quintics is,

Table[{c,Recognize[N[Sum[E^(2Pi I c^k/p),{k,1,(p-1)/5}],50],5,x]},{c,p/2}]

for prime $p\equiv1\pmod{10}$. Inspecting the resulting table of candidate quintics, identical ones with small coefficients will stand out and which gives the correct choice of $c$.

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[Math] Intuitive reasoning why are quintics unsolvable

Number Theory – Are There More Cyclic Quintics?

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