216 points by luu on July 3, 2017 | hide | past | favorite | 62 comments

 Around 1994 my dad bought this huge "Solving the Quintic" poster from Wolfram Research after he had started playing around with Mathematica. Staring at it grew a pretty good fascination in mathematics for me.I just looked into it and it turns out you can still buy it [1]! Oh, and here's a 1994 announcement about it on sci.math.symbolic [2]!
 Off-topic, but can someone explain to me how mailing lists and threads from 1994 end up on Google Groups? I never really understood what goes on there...
 Not mailing lists. These are Usenet posts. Years ago Google bought a service that was archiving the Usenet network and integrated their data into Google Groups.https://en.wikipedia.org/wiki/Usenet
 wiredfool on July 3, 2017 They bought DejaNews, who grabbed a bunch of old net.news archives from various places. They then shoehorned it into the Groups interface.
 Ordering the poster to Israel: poster \$10, shipping \$52. Wat.
 just today cleaning out my desk i found this againhttp://www.math.jhu.edu/~smahanta/Teaching/Spring10/Stillwel...>The aim of this paper is to prove the unsolvability by radicals of the quintic (in fact of the general nth degree equation for n >= 5) using just the fundamentals of groups, rings and fields from a standard first course in algebra.essentially all of the fundamentals of galois theory in 5 short (very readable!) pages.
 Another lovely aspect to this problem is that the quintic _is_ solvable if you allow yourself to use certain special functions naturally associated to the icosahedron.The genesis of the relationship is that the group of rotations of the icosahedron is the same group, A_5.
 It's solvable in many other ways. There's no rule in algebra that states each individual variable must be a different number. Thus, every variable = 0 then the equation is solved.
 No that wouldn't work. Consider the equation x^5 + 1 = 0. Clearly that isn't solvable by x=0.
 jacksnipe on July 3, 2017 That's not a "solution" in the algebraic sense that we're talking about.
 Folks might enjoy the lectures about this which V.I. Arnold taught to 14–16 year old students, https://www.mathcamp.org/2015/abel/abel.pdf
 As per https://news.ycombinator.com/item?id=14685892 this is also by Arnold.
 I have a master's in Math. none of it is new, but I'm so glad someone wrote this. expanding out all the key points.
 chii on July 3, 2017 wow, each increase in degree of the equation is literally an order of magnitude more complicated than before!
 Author here, in case anyone has any questions. :)
 This sentence is very confusing:> Now this formula looks right, since x_1x and x_2x are at the same coordinates as r_1r and r_2r.I assume you mean "since before they have been moved...", but by the time the reader gets to that point you've already told them to move r1 and r2 and x1 and x2 obviously aren't at the same coordinates.
 Thanks, I see why it's confusing. I'll fix it!
 I've seen this argument several times and always get lost at around the same point:> But that means that our candidate solution cannot be the quadratic formula! If it were, then x_1 and x_2 would have ended up swapped, too.Why?
 Because by definition, if we have found the right solution, then x_1 = r_1 and x_2 = r_2.
 I don't see why that's true by definition. We're looking for a function f st. f(a,b,c) = {r1, r2} iff 0 = a·r1^2 + b·r1 + c = a·r2^2 + b·r2 + c. Right? So as long as the correct two roots are produced, why is the order that they are produced in relevant?
 Because we start off with the candidate solution matching the real roots, and we continuously move the roots to swap them. At every point of the movement, the candidate solution match the real roots (including order) by assumption, and so if at the end of the movement the candidate solution returns to its initial state, i.e. it's equal to the swapped roots, then that would contradict continuity. Does that make sense?
 Well, some quintics are solvable, aren't they? (like x^5-1=0). Maybe for each particular quintic there is a different formula that solves it. You just prove that there's no formula that works in all cases when you plug the coefficients at the same place.It would be more satisfying (and more general) to exhibit a specific unsolvable quintic.
 Yes, that's correct. That's the big disadvantage to the topological method I talk about (developed by Vladimir Arnold); it can only prove stuff about the general nth-degree formulas, whereas Galois theory can prove stuff about specific polynomials (e.g., that x^5 – x – 1 is unsolvable in radicals).On the other hand, the topological method can be easily extended to prove stuff about formulas including other continuous, single-valued operations, like exponential and trigonometric functions, and Galois theory has a harder time with this.
 > That's the big disadvantage to the topological method I talk about (developed by Vladimir Arnold); it can only prove stuff about the general nth-degree formulas, whereas Galois theory can prove stuff about specific polynomials (e.g., that x^5 – x – 1 is unsolvable in radicals).Can these topological methods be generalized to arbitrary fields?
 I'm a bit out of my depth here, but I think you can generalize it to differential fields, where you want to figure out e.g. which functions have a closed-form integral. A reference (the only reference?) would be Topological Galois theory: https://www.springer.com/us/book/9783642388705
 dullcrisp on July 3, 2017 All quintics have solutions, but there is no general quintic formula.
 Of course, but in this game "solution" means "a number that can be expressed as a composition of radicals and elementary operations".
 moomin on July 3, 2017 Not true, there's only no general quintic formula that employs a specific, restricted set of operations.In practice, there is a general quintic formula. It just needs one extra operation.
 What other operation? If we don't know what it is, how do you know it needs just one?
 vog on July 3, 2017 Does that matter?If you are free to add any additional operation, this whole thing becomes meaningless. You can simply define your operation "NewOp(a0,a1,..,a4)" to something like "the smallest root of the quintic a0+a1x+...+a4x^4+x^5".(Here, "smallest" can be anything as long as it is a completely defined tie-breaker, such as: the value with the smallest real part, and among those the one with the smallest imaginary part.)
 I think it's something like p(x) = smallest positive solution of z^5 + z - x. Compare sqrt(x) = smallest positive solution of z^2 - x.But it's a _long_ time since I've done Galois theory and I can't find a decent math exchange answer for it right now, so don't treat this as gospel.
 Yeah, I think you're thinking of a Bring radical: https://en.wikipedia.org/wiki/Bring_radicalEdit: Oh, but that only lets you solve some quintics. https://news.ycombinator.com/item?id=14686886 describes the functions you need to solve all quintics.
 Thanks for the post. What are the special icosahedron related functions one commenter said could be used to solve quintics?
 I am the other commenter!The special icosahedron functions can be expressed in many ways but perhaps the most elementary _explicit_ representation is as particular Gaussian hypergeometric functions. See this repo for example: https://github.com/ocfnash/icosahedral_quintic The special functions are used here: https://github.com/ocfnash/icosahedral_quintic/blob/master/q...Geometrically these functions locally invert the (branched) covering represented by the diagram at the top of this page: http://olivernash.org/2012/02/05/on-kleins-icosahedral-solut...
 This page is an example why I gave up and still give up on advance mathematics.Maths is just the extreme example of science, where to make communication possible between its practitioners, new words encoding known facts are constantly created. Then another layer of new words with definitions based on the first layers are defined... and so on. Rapidly, we end-up with total gibberish for the non-initiated.For example on that page, we start with quintic and radicals (I can grok that). Then icosahedral functions, then hypergeometric functions, finite monodromy, 60-fold branched covering of the complex projective line (lost!).
 19eightyfour on July 3, 2017 Thanks then.
 akalin on July 3, 2017 Hey, this is really cool!
 Thanks for this, this is a topic I've been curious about for a long time, but never knew how to learn about it without diving into a textbook.I'm reading through it now, and I'm confused by this statement: "Second, and more surprisingly, if you swap r_1 and r_2r, x_1 and x_2 also exchange places, seemingly contradicting Theorem 1!"Why is there any connection to theorem 1 here? We're talking about a radical, and theorem 1 is concerned with rational expressions.
 Which library/libraries did you use for the interactive graphs?
 Just JSXGraph: https://jsxgraph.uni-bayreuth.de/wp/index.htmlI'm not entirely happy with the framerate I get with the button-triggered animations (especially on mobile) but lack the front-end expertise to track down the problem.
 Would be nice if dragging the coefficients would also change the roots.
 Yeah. I think I punted on that because, although it's easy for degrees <= 4 (since those have general formulas!), I'd have to implement some root-finding method for degrees >= 5, and somehow preserve the orderings of the roots. Not insurmountable problems -- I might go back and do that someday!
 I was just pondering about this. I imagine that moving a coefficient a small amount (and so long as the roots are not too close) that Newton-Raphson, initialized with a previously-known root, would converge very quickly to the new root while preserving the order.If there truly are roots of multiplicity >= 2, then preserving the order of the multiple roots doesn't matter. Another issue is that Newton-Raphson might fail if the roots are too close.
 TheRealPomax on July 3, 2017 Ah, interesting, I had not heard of that one.
 It doesn't work with Firefox, right?
 I just tried with Firefox 54.0.1 on macOS, and it seems to work. What version are you using?
 Is there an analogy with the way the sequence from reals to complex numbers to quaternions to octonions (iterations of the Cayley-Dickson construction) loses algebraic properties?https://en.m.wikipedia.org/wiki/Cayley%E2%80%93Dickson_const...
 Numerical methods quite easily give the roots of arbitrary polynomials of much higher order. QR-iteration works well up to some polynomial order, say at least 20. The idea is to construct a matrix that has the studied polynomial as its characteristic polynomial, and find the eigenvalues using a repeated QR decomposition. This gives you all complex and real roots.
 Nobody was saying otherwise.
 The parent comment wasn't saying that nobody was saying otherwise...
 Interestingly it was first proved by Abel when he was 22 years old.
 Link to cached version since original is down: http://cc.bingj.com/cache.aspx?q=why+is+the+quintic+unsolvab...
 Wow. People still use Bing?
 I recently (couple months ago) switched to bing and edge, I am fed up of google's abuse of power. I must say I don't find myself constrained.
 Bing rewards baby
 Walf on July 3, 2017 "Good baby; have some more SERPs."
 Very nice! The covering transforms have similar structure to the Galois theory. It is cool to see them in action visually.
 If you extend the operations allowed it's possible using some exotic functions: https://en.wikipedia.org/wiki/Thomae%27s_formula#cite_note-t...
 he does a pretty good job of connecting unsolvability by radicals to the unsolvability of the permutation group S5.I suspect what happens these quintics define Riemann surfaces and perhaps the premutation of the roots an be mapped unto the fundamental group of the surface.
 Modulo 5 it is the full transformation semigroup on 5 elements.
 Very nice, Thank you Always wanted to know, now you saved me a year dedicated to abstract algebra

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