The unsolvability of a general quintic equation in terms of the basic arithmetic operations and $n$th roots (i.e. the Abel–Ruffini theorem) is considered a major result in the mathematical canon. I have recently become confused as to why this is the case.

The formula $z=\frac{-b+\sqrt{b^2-4ac}}{2a}$ expresses the solutions to the quadratic equation $az^2+bz+c = 0$ in terms of the inverse of an analytic function $z \mapsto z^2$. We have simply turned the problem of inverting one analytic function, $z \mapsto az^2+bz$, into the problem of inverting another analytic function, $z \mapsto z^2$. Therefore, all the power of the quadratic equation lies in how it solves any quadratic equation by inverting a single analytic function, $z \mapsto z^2$.

Similarly, Cardano's formula solves any cubic equation by inverting a two analytic functions, $z \mapsto z^2$ and $z \mapsto z^3$. Interestingly, you can also solve a cubic by inverting only one analytic function, for example $z \mapsto \sin z$.

And crucially, you can also do this for quintic equations, by inverting $z \mapsto z^k, k \leq 4$ and $z \mapsto z^5+z$.

One possible statement of the Abel–Ruffini theorem is that it is impossible to solve a general quintic equation by exclusively inverting functions of the form $z \mapsto z^k$ for $k \in \mathbb{N}$. But why would we only be interested in solutions that invert analytic functions of that form? In simplier terms, what's so special about radicals that makes solutions in terms of them so desirable? I can't see an argument that such inverses are intuitively straightforward: they often produce answers that purely formal (e.g. $\sqrt{2}$ doesn't have a simpler defintion than the positive inverse of the squaring function at $2$).

To me, it seems that the more natural question is, for $n \in \mathbb{N}$,

Is there a finite set of analytic functions such that the solutions to any degree $n$ polynomial may be expressed in terms of the inverse of these analytic functions?

I know very little about the status of this question (exept that it holds for some small values of $n$). Any information on what is known about this question would also be of interest.

  • 53
    $\begingroup$ The only reason mathematicians nowadays make such a big deal about it is that it was a very fruitful problem, i.e., the techniques developed to solve it have been helpful for many other problems. $\endgroup$
    – Will Sawin
    Commented Jan 9, 2022 at 16:25
  • 29
    $\begingroup$ Radicals were not preferred because of the corresponding field extensions. The idea of a radical formula just came out of the discovery of root formulas in low degree. That there is a translation of the concept into the language of field extensions is simply part of the process of explaining how to turn the existence of a radical formula into the existence of a certain structure in a Galois group using today’s interpretation of Galois theory. Nobody was thinking about field extensions 200 years ago. $\endgroup$
    – KConrad
    Commented Jan 9, 2022 at 17:17
  • 74
    $\begingroup$ Literally every algebra text: "Galois theory is cool because it shows there is no formula for the quintic." Passer-by: "I wonder why this insolubility of the quintic thing is made such a big deal of." Mathematicians in reply: "Who is making a big deal of it? No one thinks it's a big deal. How very strange that you somehow got that impression." $\endgroup$
    – R.P.
    Commented Jan 9, 2022 at 19:39
  • 94
    $\begingroup$ A lot of y'all are jaded! When I was 18, I had been solving polynomials for years and felt it was pretty natural to wonder or not there was some kind of equation for their solutions. And the idea that we could not only state this question precisely, but even answer it was pretty mind-expanding. The fact that Galois theory is a broader context that is useful elsewhere was just icing in the cake. I've moved onto harder (mathematical) drugs by now, but surely "answers a classical question that many students will have at least implicitly thought about" is a good reason to teach something... $\endgroup$ Commented Jan 9, 2022 at 23:23
  • 59
    $\begingroup$ (in other words: I'm very skeptical of the idea that we should design our undergraduate curriculum arounds our current research interests or what is "hot" right now. Also, however lame it might make me sound I hope I never completely forget the childlike wonder of this style of mathematics.) $\endgroup$ Commented Jan 9, 2022 at 23:24

12 Answers 12


I think that a large part of the difficulty we have in understanding why this result is considered important is that it is psychologically difficult to put oneself into the shoes of mathematicians of the past. There was a time not so many centuries ago when people didn't know how to solve cubic equations with radicals. Whether the quintic is solvable in radicals was once a difficult question. A problem that gains some notoriety for being difficult is usually going to be considered important when it is finally solved, regardless of whether it ends up occupying a central place in the "theory" that we end up constructing a posteriori. Fermat's Last Theorem is another good example of this.

Occasionally I will hear mathematicians say something to the effect of, "Fermat's Last Theorem isn't important; it's the math used to prove the theorem that is important." I don't entirely agree. There are two different kinds of importance that are being conflated. Something can be important because it occupies a central position in our theory. But an appealing and tantalizingly difficult problem is important because of its role in capturing our imagination and giving us something to sink our teeth into. I do not think we should disavow the importance of such problems just because they are solved. Many of our much-beloved theories would likely not exist if there hadn't been some interesting unanswered questions to motivate our research.

The quintic has the additional feature of being an "impossibility" result. Like non-Euclidean geometries and Gödel's incompleteness theorems, the solution made us realize that we had been making some unfounded assumptions about what the answer should look like. The psychological broadening of our horizons was a valuable byproduct that is sometimes overlooked.

  • 7
    $\begingroup$ I really like this answer and think it makes some great points. Although I worry about how far the great cultural importance we place on certain results can take us from the actual math. For me, it's a bit shocking to realize that, despite 'the unsolvability of the quintic' being a big part of the mathematical landscape (esp. w.r.t. outreach) there is no deep mathematical sense in which quintics are unsolvable. $\endgroup$
    – Arthur
    Commented Jan 9, 2022 at 18:14
  • 4
    $\begingroup$ @Arthur The sense of "solvability" here is not devoid of mathematical interest. After all, solvable groups play a very important role in group theory. You might also enjoy my paper, What is a closed-form number? In particular, Corollary 1 shows that "solvable in radicals" (probably) coincides with a somewhat more general definition of solvability. $\endgroup$ Commented Jan 9, 2022 at 21:33
  • 1
    $\begingroup$ @TimothyChow Thank you for pointing me towards your paper, I enjoyed reading it. The content is certainly relevant, and Corollary 1 does (conjecturally) provide a surprising answer to my original question. My new question would then be, why do we talk so much of about certain quintic equations being unsolvable and so little about $\mathbb{E}$? From your paper, it sounds like you certainly agree with the second half of my new question. $\endgroup$
    – Arthur
    Commented Jan 9, 2022 at 22:52
  • 1
    $\begingroup$ @Arthur Well, it turns out to be hard to prove much rigorously about $\mathbb E$. Skimming the other answers and comments, I see only one brief mention of hypergeometric series. You may be interested in the paper by Sturmfels, Solving algebraic equations in terms of $\mathscr{A}$-hypergeometric series, which partially addresses the latter part of your question. $\endgroup$ Commented Jan 10, 2022 at 17:21
  • 1
    $\begingroup$ "The psychological broadening of our horizons" -- but it's not psychological, it is mathematical broadening. $\endgroup$
    – Wlod AA
    Commented Feb 23, 2023 at 6:35

It is not a big deal anymore (edit: I mean in modern math, not for teaching undergraduate math courses) and has not been for a long time. It was important historically because of the math that came out of efforts to solve it, namely many important concepts in group theory (permutation groups, normal subgroups, quotient groups) and Galois theory. Solvability of polynomials in radicals is often the culmination of a course on Galois theory but that result in no way reflects why Galois theory is important in contemporary math (which it is).

Galois theory in turn was very important for math even outside of algebra, e.g., the Galois correspondence inspired Lie to study Lie groups (to create a Galois theory for differential equations even if that is not how Lie theory is used today) and we can see parallels to it in the correspondence between covering spaces in topology and subgroups of the fundamental group.

Your emphasis on Abel-Ruffini and inverting analytic functions suggests you may have missed the point of what Galois did compared to Abel and Ruffini. The Abel-Ruffini theorem was about “generic polynomials” and was purely a negative result. Their work could not be applied to specific polynomials with rational coefficients, like $x^5-x-1$. Galois theory, in contrast, could be used on specific polynomials and works over fields that need not be inside the complex numbers.

By the way, Galois himself was pessimistic about the idea of actually applying his work to decide if a specific polynomial you did not already know something about is solvable by radicals. See pp. 225-231 of https://uberty.org/wp-content/uploads/2015/11/Peter_M._Neumann_The_Mathematical_Writings.pdf; the English translation is on pp. 227 and 229. The key part:

If you now give me an equation that you have chosen at will, and you wish to know whether or not it is solvable by radicals, I will have nothing to do other than to indicate to you the way to respond to your question, without wishing to charge either myself or anyone else with doing it. In a word, the calculations are impractical. [...] But most of the time in applications [...] one is led to equations all of whose properties one knows beforehand: properties by means of which it will always be easy to answer the question by the rules which we shall expound.

Even though taking roots is not enough to solve polynomial equations in algebra, inverting $z \mapsto z^k$ does suffice locally for maps between Riemann surfaces (created by Riemann 20 years after Galois) since every nonconstant holomorphic map between Riemann surfaces looks near each point like $z^k$ for a unique $k$ in suitable coordinates. So the phenomenon of ramification for such maps is based on the adequacy of power functions $z^k$ as a local model formula in single-variable complex analysis.

  • 27
    $\begingroup$ Or to paraphrase Kermit the Frog, "Someday we'll find it, the Galois connection // the covers, the scheme-rs and me" $\endgroup$ Commented Jan 9, 2022 at 17:18
  • 1
    $\begingroup$ According to Wikipedia, the name 'Abel–Ruffini theorem' is used to denote both the result about “generic polynomials” and the stronger result that there are specific equations of degree five and higher that cannot be solved by radicals. Perhaps using that term was needlessly ambiguous. $\endgroup$
    – Arthur
    Commented Jan 9, 2022 at 17:26
  • 11
    $\begingroup$ It seems you discovered a mistake on the internet. I would not call the stronger result the Abel-Ruffini theorem. When I looked in 10+ algebra books that discuss solvability by radicals, none referred to the stronger result as Abel-Ruffini. And when they showed the generic polynomial of degree $n$ is not solvable by radicals, they all made a point that this did not apply to specific polynomials in $\mathbf Q[x]$, for which they'd need the work of Galois. At the same time, naming a theorem after someone who only worked on some special case of it is pretty common, e.g., Lagrange's theorem. $\endgroup$
    – KConrad
    Commented Jan 9, 2022 at 18:06

Perhaps it's not a big deal that's it's true; but it's a big deal that we can know and prove that it's true. I don't mean because the ability to prove that it's true implies the ability to do other, more useful things; I mean because it's not a kind of statement that seems like it could be proven true.

I distinctly recall learning as a young proto-mathematician that there was no way to double a cube or trisect an angle using straightedge and compass, no general solution to the quintic using radicals, and no closed-form integral of $\mathrm{e}^{-x^2}$, and I found it absolutely mind-boggling that these things could be true beyond "well, we haven't found a way to do it and we've looked really hard". This was before there were good resources for this kind of thing available on the internet, so to be honest, I still thought they might just be rumours that had been distorted in the retelling.

So as a high-school student it was hard to imagine that this is even a type of thing that can be known. A few years later I was an undergraduate at the University of Cambridge studying Galois Theory, and the lecture in which the insolubility of the quintic was proved (as a corollary of a much more general theorem) certainly felt like a milestone: look how far you have come.

  • 3
    $\begingroup$ I had the same response to the question. The Abel-Ruffini theorem is not particularly significant, in and of itself, from the perspective of higher mathematics. But from a student’s perspective, it is surprising and significant. After learning the quadratic formula, and that cubics were also solvable, I implicitly assumed that there was some equivalent for higher order polynomials involving higher order roots. I think most of my classmates did as well. I have mentioned this result to a bright friend who studied engineering and he was shocked by it, since he had never encountered it before. $\endgroup$
    – sasquires
    Commented Jan 12, 2022 at 19:41
  • 2
    $\begingroup$ @sasquires yeah, as an engineer, the problems/proofs I view with alternating awe and suspicion are when going from e.g. an exponent/dimension/etc of 4 to 5 changes the result from "here's an easy answer" to "there's provably no solution"/"20 years of work hasn't let us figure it out yet". 0, 1, 2, I can grudgingly accept as special cases. How does going from 4 to 5 suddenly mean we need whole new fields of math to answer the question? Why did it stop generalizing in such a way? I'm happy there's smart math people to handle it for me. $\endgroup$
    – mbrig
    Commented Jan 13, 2022 at 5:01
  • 2
    $\begingroup$ Many mathematicians think, "There's no reason for engineers and others who aren't professional mathematicians to care about the 'unsolvability' of the quintic, because in the end, numerical answers are what are needed, and numerical methods work fine for polynomial equations of any degree." Ironically, in my (anecdotal) experience, the people who seem most intrigued by the lack of a "closed-form" solution to the quintic are engineers and others who aren't professional mathematicians. $\endgroup$ Commented Jan 13, 2022 at 9:47
  • 2
    $\begingroup$ @mbrig That's a really good point, results are much more intriguing when the boundary between "solvable" and "not solvable" is some larger integer like 5. For the record, the relevant difference between 4 and 5 here is that A_4 (the alternating group on 4 elements) is not simple but A_5 is. $\endgroup$
    – kaya3
    Commented Jan 13, 2022 at 10:07

Mathematically, there is nothing special about radicals: solving algebraic equations by radicals is only one example of mathematical problems among others that are self-contained, comprehensive, complete and decidable in a given class of functions/expressions/numbers.

The question of solving algebraic equations by radicals comes from the historical question of solving algebraic equations by straightedge and compass. Today, this mathematical problem introduces Galois theory.

But there is also another branch: Topological Galois theory of Arnold and Khovanskii.

[Khovanskii 2021]:
"As was discovered by Camille Jordan the monodromy group of an algebraic function is isomorphic to the Galois group of the associated extension of the field of rational functions. Therefore the monodromy group is responsible for the representability of an algebraic function by radicals ... .
Theorem 12. If the monodromy group of an algebraic function is unsolvable then one can not represent it by a formula which involves meromorphic functions and elementary functions and uses integration, composition and meromorphic operations."

Let $\mathbb{E}$ be the explicit elementary numbers of [Chow 1999].

[Chow 1999]:
"Corollary 1. If Schanuel's conjecture is true, then the algebraic numbers in $\mathbb{E}$ are precisely the roots of polynomial equations with integer coeficients that are solvable in radicals."

That means: An algebraic equation that cannot be solved by radicals can neither be solved by explicit elementary numbers / in terms of explicit elementary functions nor by the other functions mentioned in Khovanskii's theorem 12 above.
$\ $

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory - Solvability and Unsolvability of Equations in Finite Terms. Springer 2014

[Khovanskii 2019] Khovanskii, A.: One dimensional topological Galois theory. 2019

[Khovanskii 2021] Topological Galois Theory - Slides 2021, University Toronto


To understand what is "so special about radicals that makes solutions of algebraic equations in terms of them so desirable" we should look at the problem not from the point of view of modern higher mathematics but from the point of view of mathematics as it was conceived possibly up to the fifties of the 20th century. Up to that time, there was not a clean distinction between pure and applied mathematics: mathematics was conceived as the "queen and servant to the other sciences". And in this context, asking

"Can I solve this algebraic equation in term of radicals?"

equals to ask

"Can I construct these intersections by using straightedge and compass (or just by using compass, after Georg Mohr and Lorenzo Mascheroni)?".

And knowing that there are some "elementary" constructions which I cannot do with such basic and ubiquitously used tools (not only in "R&D" offices but in every single mechanical workshop) means knowing that there are designs of some mechanical parts which are incomparably more difficult to achieve (in an exact way) respect to other, apparently similar, ones.
Thus, from the applied side of mathematics, it has been incomparably important to figure out this intrinsic limitation in the tools used, as this also in turn has forced the savants to develop new methods to design more and more complex mechanical parts: after all, we could also remember that Lorenzo Mascheroni, military engineer of Napoleone Bonaparte, developed his "compass" geometry in order to reduce the errors due to the use of a straightedge involved in making geometric constructions .

  • 4
    $\begingroup$ This would capture the radicals better if you said “straightedge and slide rule”. $\endgroup$
    – user44143
    Commented Jan 9, 2022 at 22:35
  • 1
    $\begingroup$ @MattF., a hilarious phrase! :) $\endgroup$ Commented Jan 9, 2022 at 23:16
  • 8
    $\begingroup$ Yes, this answer is a little misleading, since constructibility using straightedge and compass is equivalent to being algebraic of degree a power of two (over the base field). This isn't the same as being solvable in radicals. For example, duplication of the cube is impossible with straightedge and compass, but $2^{1/3}$ is obviously expressible using the cube root operation. $\endgroup$ Commented Jan 10, 2022 at 1:30
  • 8
    $\begingroup$ @TimothyChow constructible implies algebraic of $2$-power degree over the base field, but not conversely: the field the number generates must be in a tower of successive quadratic extensions, and extensions of degree $4$ might not have a quadratic subextension. For example, $f(x) = x^4 + 8x + 12$ is irreducible over $\mathbf Q$ and its splitting field over $\mathbf Q$ has Galois group $A_4$. Since $A_4$ has no subgroup of index $2$, if $\alpha$ is a root of $f(x)$ then $[\mathbf Q(\alpha):\mathbf Q] = 4$ and there's no quadratic field in $\mathbf Q(\alpha)$, so $\alpha$ is not constructible. $\endgroup$
    – KConrad
    Commented Jan 10, 2022 at 3:10
  • 2
    $\begingroup$ @TimothyChow IOW, a number is constructible iff it is contained in a Galois extension whose degree is a power of 2. $\endgroup$ Commented Jan 10, 2022 at 9:40

The question of solvability by radicals led naturally to the study of solvable groups, and solvable groups play an important role in many parts of algebra and number theory, being in some sense the easiest groups after abelian groups. Just as one example, Wiles's proof of the modularity conjecture, and thus of Fermat's Last Theorem, relies on the fact that a certain small (non-abelian) matrix group is solvable, since that allows him to apply a theorem of Jerry Tunnell's on properties of Artin $L$-functions for Galois extensions of $\mathbb Q$ that have a solvable Galois group. So you could rephrase this by saying that Wiles' proof works because certain equations (roughly, that define the points of order 3 on the elliptic curve) are solvable by radicals. So this is an example of solvability by radicals arising in an unexpected place. And if, for example, Wiles needed to use the points of order 7, then (if I understand correctly) the proof would have fallen apart because we don't yet know enough about $L$-functions for fields that are not generated by radicals.


There are already some great philosophical reasons here, so let me give a very down-to-earth practical reason.

Suppose you have no electronic devices, but you have a great big book of logarithms. If you want to find the square root (or fifth root or whatever) of a number, you can do that in a few straightforward steps - look up the logarithm of the number, divide it by 2, and then do the reverse lookup to find which number has that as its logarithm.

What would be the quickest way to solve a cubic with this setup? Depending on the required precision, it could very well be to use the cubic formula! Of course there's Newton's method or other iterative methods, but the advantage of using the book of logarithms is you only need to do the calculation once and you already have the max precision.

For a general quintic there's no way to use the book of logarithms this way - you have to resort to iterative methods. I think that's noteworthy.

  • 3
    $\begingroup$ How do you find the third root of a complex number with a book of logarithms? $\endgroup$
    – abx
    Commented Jan 10, 2022 at 20:30
  • 2
    $\begingroup$ In polar coordinates presumably $\endgroup$ Commented Jan 10, 2022 at 20:40
  • $\begingroup$ I'm interested to know what method(s) the mathematicians used in the past to find roots. They certainly knew better than us about practical efficiency, with "tricks" to speed-up the computations. I would guess that in general Newton's method is more efficient (and it allows higher precision). $\endgroup$ Commented Jan 10, 2022 at 21:36
  • 8
    $\begingroup$ Fortunately, it seems that good published mathematical tables included not only logarithms but also trigonometric functions :) $\endgroup$ Commented Jan 10, 2022 at 21:58
  • 3
    $\begingroup$ @MattF. huh? Trisection of an angle in polar coordinates is simply division by three. The part that does require some extra effort is transforming to polar coordinates in the first place, but again tables (sine and arctangent) help. $\endgroup$ Commented Jan 12, 2022 at 11:26

The interest of algebraic solutions by radicals is a little similar to the interest of geometric solutions by ruler and compass. Part of the charm is that, while it is straightforward (in many cases) to see what can be done, it is mysterious how one would ever show something can't be done -- whether that something is squaring the circle, or solving a quintic polynomial.

As for radicals, it is clear from Klein's work that solving x^n =a was considered "natural and explicit" because one can find the roots using logarithms. (Set x = exp((1/n) log a).
The use of transcendental functions here was not at all off-limits, in fact it was one of the main ideas. A "formula" for the roots of a polynomial involving radicals is therefore regarded as an "explicit" (if multivalued) function built using the elementary operations of +/-, *, division, and x -> x^(1/n). Obviously something new has to be added, and the radical operation is one of the simplest after the arithmetic operations.

In his lectures on the icosahedron, Klein similarly has no reservations about using quite sophisticated modular functions just to compute the inverse of a specific degree 60 rational map, which can in turn to be used to solve quintic equations.

In modern terms, one could argue that solving the equation x^n = a is distinguished and routine since Newton's method is highly reliable and rapid in this case. Indeed for a>0, any real initial guess x>0 will converge and the number of digits of accuracy will double with each iteration. For complex a, any initial guess chosen at random will also converge (i.e. the Julia set has measure zero, indeed Hausdorff dimension < 2).


For me there's a very simple and important answer, beyond what has been said already about the historical value of the work that was motivated by the problem.

And that answer is that the problem is natural and inherently interesting.

Why is any pure maths done? Why do we pay particular attention to some questions? Why, for example, do we care about the twin primes conjecture? It's not for any practical reason. If you mess around with numbers, it won't be long before you discover the concept of primes, and a while later you will notice twin primes and wonder how many of them there are. The question arises naturally as a product of observation and curiosity.

You may as well ask why we care so much about primes. "Sure", you say, "we can factor numbers into primes, but we can also factor numbers into things that aren't primes, so who cares?". The answer is that we find primes inherently interesting because of their simplicity (having the least possible number of divisors).

It's the same with roots and solving polynomial equations. If you play around with algebra, you'll naturally discover the concept of roots. You'll discover that low-degree polynomials can be solved using roots and nothing more, and you'll naturally ask whether this is true of all polynomials. Sure, you can also say that the solutions to $x^2 -2x -1 = 0$ are "whatever the solutions to $x^2 -2x -1 = 0$ are", but we have a special interest in solutions of the form $x^2 = a$ because of their simplicity, and so we find it interesting to learn that these are all that is required.


Rephrasing in part some of the previous comments and answers, my take is that historically roots, like powers, were seen not as "functions" in the modern sense, but as natural "operations" extending the four standard ones. Then naturally mathematicians became curious about whether the resulting enlarged family of numbers (or "field" in modern language) would

A) consist always of solutions of polynomials;

B) contain all such solutions.

Solving the cubic was perhaps the major breakthrough in exploring B), then trying to solve the quintic became the inevitable stumbling block.

This remains a fascinating story because the historical path is after all not different from the natural development of many young adults' mathematical curiosity. As pointed out by others, we tend to forget it as our mathematical sophistication grows.

As an aside, I'd be curious to know when A) was settled historically. I'd be surprised if it hadn't been known at least by Lagrange or Euler, but then I'm not quite sure if rigorous notions of field extensions, degree, linear dependence were all in place then.

  • $\begingroup$ For part A, you're just asking whether an explicit expression involving radicals is algebraic, right? This doesn't require any notion of field extensions or degrees or linear dependence; you just need an algorithm that starts with something like $x = \sqrt{3} + \sqrt{5}$ and eliminates all the radicals by repeatedly taking powers and shuffling things around. This might have been considered "obvious" by someone like Euler, and not proved formally. $\endgroup$ Commented Feb 24, 2023 at 19:06
  • $\begingroup$ @TimothyChow You are correct as to what I meant. I don't understand what type of algorithm you have in mind. How would it handle something like the sum of 6 square roots, or 4 cube roots, or something even more complicated? Noticing that the field generated by the sum or product of two algebraic numbers is also a finite dimensional vs over $\mathbb{Q}$ is only trivial if one knows the basic properties of fields and vector spaces, I think. (I don't mention ratios since $a/b$ is just $a\times 1/b$.) $\endgroup$ Commented Feb 25, 2023 at 21:19
  • $\begingroup$ For different proofs see math.stackexchange.com/questions/331017/… and other links there. But no mention of the history of this result. $\endgroup$ Commented Feb 25, 2023 at 23:22
  • $\begingroup$ If you have a rational linear combination of $\sqrt{p_1},\ldots, \sqrt{p_n}$ with $p_i$ prime then repeated squaring and rearranging will always give you a rational linear combination of terms of the form $\sqrt{b}$ where $b$ is a (squarefree) product of some subset of the $p_i$. So you can always make progress by isolating the smallest $\sqrt{b}$ in your current expression on one side of the equation, and then squaring; once gone, it will never come back. I admit I have never formally fleshed out this idea and just assumed it would be straightforward. $\endgroup$ Commented Feb 25, 2023 at 23:23
  • 1
    $\begingroup$ The result you linked to is harder since you're dealing with algebraic numbers that are not expressible in terms of radicals. I think people would sense the need for a formal proof of that one, so maybe one could track down the history. $\endgroup$ Commented Feb 25, 2023 at 23:24

So many answers claiming that solving polynomials in radicals (read: finding their roots quickly) is not important! Here's my (and many others' who worked on graphics apps) train of thought at one time or another:

"Bézier curves make transitions very smooth because the first derivatives of adjacent sections coincide. Let's make the app even smoother with quintic curves so that the second derivatives coincide too!"

"Oh, wait, that would mean that I'd have to solve thousands of quintics per second, and I don't have a closed form solution for that! I would have to run thousands of interpolations rather than closed formula substitutions! Screw that; let's keep cubics."

  • 10
    $\begingroup$ Solving algebraic equations in radicals is not useful for numerically computing their roots. Roots of algebraic equations are numerically computed by a variant of Newton's method, not by using closed-form expressions in radicals even when these are known, except in degree $2$ or for other very special forms. Indeed, $n$-th roots themselves are typically computed numerically using Newton's method, so they are not really any simpler than general algebraic equations. $\endgroup$
    – Gro-Tsen
    Commented May 10, 2022 at 0:23
  • 4
    $\begingroup$ @Gro-Tsen, $n$-th roots computations are done in hardware; IIRC with CORDIC rather than Newton's method, and that's a lot faster than iterative methods. I think the reason for higher degree algebraic equations being solved with Newton's method is precisely because there are no closed form formulae to solve them with radicals. $\endgroup$
    – Michael
    Commented May 10, 2022 at 15:54
  • 3
    $\begingroup$ @Gro-Tsen. Solving equations by using formulas involving radicals certainly IS useful, in my experience. For cubics, it’s typically faster than coding your own iterative algorithm. There are potential problems with overflow and subtractive cancellation, but these are surmountable. The algebraic solutions might need to be polished using a Newton-Raphson step or similar, but, at the very least, they provide an excellent place to start this polishing, I agree with Michael — sometimes decisions about software design are dictated by solvability via radicals. $\endgroup$
    – bubba
    Commented Sep 19, 2022 at 11:52
  • 2
    $\begingroup$ @Michael I'm still having trouble believing that closed forms are relevant here. Closed form or not, hardware or software, CORDIC or NR, solving quintics is going to be slower than solving cubics, so if speed is paramount then quintics will be rejected regardless. Let me ask this. Quartics are solvable in radicals. Does anybody actually use the "closed form" solution for quartics to solve quartics in practice? It's a real mess of a formula; extracting 4th roots isn't the only tricky part. $\endgroup$ Commented Feb 23, 2023 at 13:26
  • 3
    $\begingroup$ By the way, even if you have a cubic equation and are interested only in real solutions, the closed-form formula requires you to extract complex cube roots and not just real cube roots, so it still doesn't trivialize the problem even if you have a hardware cube root. $\endgroup$ Commented Feb 23, 2023 at 13:32

I like this question because I agree with its sentiment. Let me give an additional reason why the insolubility of the quintic is an overrated result in my opinion. I believe that we shouldn't even be saying that the general cubic is solvable in radicals. The reason being that we can't solve the general cubic if we restrict ourselves to the real numbers entirely, which I find an very reasonable restriction (see here). In my entire mathematical education, I've heard it mentioned only once that you need complex numbers to solve the general cubic, even though the roots may be real, which I find very surprising. Worse still, the supposed "solubility of the cubic" is widely advertised even in texts and presentations which are obviously aimed at audiences that are not necessarily familiar with complex numbers.

  • 4
    $\begingroup$ In fact, not only are complex numbers required to solve the cubic, that's why complex numbers were invented in the first place! Whether or not one should call that "solving" the cubic, certainly the invention of complex numbers is a big deal in the history of math. As for mathematical education, there is now a Veritasium video which explains all this and which has 14 million views as of this writing. $\endgroup$ Commented Feb 23, 2023 at 13:00
  • $\begingroup$ It is a good video, but it does not mention the fact that you actually need complex numbers to solve the general cubic equation. Nothing at all that hints at the impossibility result mentioned on the Wikipedia page that I linked to. So it seems that the video is quite irrelevant w.r.t. the point I was making. $\endgroup$
    – R.P.
    Commented Feb 24, 2023 at 12:08
  • 1
    $\begingroup$ It's true that Veritasium doesn't explicitly mention the result of Wantzel, whose role in impossibility results (duplication of the cube, trisection of the angle) is often underappreciated. But I think the video does hint at the result. It does mention that Cardano's formulas don't let you solve the cubic if you restrict to the real numbers, and that this fact was what led people to seriously consider imaginary numbers. That is a hint, even though it's of course technically possible that all those people missed some other formula which allows one to bypass complex numbers. $\endgroup$ Commented Feb 24, 2023 at 13:04
  • $\begingroup$ The worst thing about not mentioning the need for complex numbers to solve cubics, even those with real roots, is that it leads people with the wrong impression that complex numbers were invented to solve quadratic equations. $\endgroup$
    – LSpice
    Commented Feb 24, 2023 at 14:12
  • $\begingroup$ @TimothyChow Thanks for the reference to Wantzel, I was not aware that he was also the discoverer of this proof. To quote from his 1843 article: "Ce travail a pour but de montrer que [...], et que, par la résolution des équations numériques de degré supérieur au second, on obtient de nouvelles quantités irrationnelles qui ne peuvent s'exprimer par des radicaux." So either Wantzel was over-selling his result here, or it would seem that "solvable in radicals" really meant "solvable in real radicals" back when these discoveries were actually being made. $\endgroup$
    – R.P.
    Commented Feb 24, 2023 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.