In this document by Jim Brown it is claimed (on Section 3, pg 5) that:

[Descartes] believed that all polynomials of degree $>4$ could be solved with the same methods as had been applied to the quadratic, the cubic, and the quartic. In fact, he left the solution of higher degree equations as an exercise to the reader.

It is a well known fact that the general quintic, and all higher degree polynomials, can not be completely resolved in terms of radicals and elementary operations. So, such an exercise would be impossible to complete, making this an interesting historical example of mathematical hubris.

However, it's not clear where the author is getting this information from, and I couldn't find any first-hand sources for this.

Is it true that Descartes left solving the quintic, and higher degree polynomials as an exercise to his readers? If so, could I get a specific reference for this?

Edit: I contacted the author of the above document, and he replied saying he doesn't recall where he got this information.

  • 3
    $\begingroup$ Why not send him an email and ask him? It's not as if he's a 19th century mathematician who died 90 years ago . . . $\endgroup$ Commented Aug 6, 2019 at 9:39
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    $\begingroup$ @DaveLRenfro I contacted the author, he says that he doesn't recall where he got that information from. $\endgroup$
    – ZKG
    Commented Aug 6, 2019 at 15:55
  • $\begingroup$ You should probably include this new information (the author doesn't remember) in your question. $\endgroup$ Commented Aug 6, 2019 at 16:23
  • $\begingroup$ @DaveLRenfro Good point, I've updated the post. $\endgroup$
    – ZKG
    Commented Aug 6, 2019 at 16:46
  • $\begingroup$ Certainly not "an interesting exemple of mathematical hubris". Descartes believed his method for solving Pappus' problem to be universally valid and that solving cases with n>4 would show how to proceed with polynomials. All this is in his Geometrie and a reference is most certainly to be found there. Without a precise wording one has to look for a paraphrase through the whole text (which isnot long). $\endgroup$
    – sand1
    Commented Aug 6, 2019 at 17:38

2 Answers 2


It is in the third book of La Géometrie:

I could also add rules for equations of the fifth, sixth, and higher degrees, but I prefer to consider them all together and to state the following general rule :

...and, consequently, if it is of the third or fourth degree, the problem depending upon it is solid; if of the fifth or sixth, the problem is one degree more complex, and so on. I have also omitted here the demonstration of most of my state- ments, because they seem to me so easy that if you take the trouble to examine them systematically the demonstrations will present themselves to you and it will be of much more value to you to learn them in that way than by reading them. -- Dover Publ., p. 192 = Archive.org, p. 192.

One question that arises for me is what a "solution" is for Descartes. I'm just not familiar enough with Descartes to say authoritatively. He is at some pains to argue that one ought to "admit" into geometry curves beyond the circle and conic sections in solving problems. In the section quoted above, he is discussing roots of equations, which may be interpreted as points of intersection. However, the solutions tend to be given in algebraic form, although many of the examples are geometric problems and the solutions are effected by the construction of curves. He shows, as an example, how to solve a degree 6 problem (intersection of a circle and a cubic), and he gives a method to construct the cubic ("In this way we can find as many points of the curve as may be desired", p. 228). But it is clear from the discussion that when you actually draw the curves, you can run into difficulties:

It should be remarked, however, that in many of these problems it may happen that the circle cuts the parabola of the second class so obliquely that it is hard to determine the exact point of intersection. In such cases this construc- tion is not of practical value. (p. 239)

He sums up in the conclusion by asserting the generality of his method for solving all problems, the problems being, I take it, geometrical:

...furthermore, having constructed all plane problems by the cutting of a circle by a straight line, and all solid problems by the cutting of a circle by a parabola ; and, finally, all that are but one degree more complex by cutting a circle by a curve but one degree higher than the parabola, it is only necessary to follow the same general method to construct all problems, more and more complex, ad infinitum... (p. 240)

My reading is that Descartes is not approaching the solution of problems with the same restrictions in mind that "solution by radicals" entails. But my reading might be too cursory.

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    $\begingroup$ I haven't tried looking at (an English translation of) Descartes' book, but from what you've provided, it certainly seems to me that he is simply talking about solving by graphical methods (which would have been fairly novel at the time), without worrying about the kinds of things that were later considered important. Perhaps the only thing being white-washed is the issue of whether graphs of polynomials of arbitrarily high specified degrees exist as curves that can be used in the same way as conics, cubics, etc. (e.g. not transcendental or "mechanical curves"). $\endgroup$ Commented Aug 7, 2019 at 18:48
  • $\begingroup$ I ran into the same problem as well; in La Géometrie Descartes does not seem to explicitly claim that higher degree polynomials can be solved in terms of radicals. He seems more concerned with geometrical methods, so it's not completely clear what he would consider to be an acceptable "solution." That being said, his methods definitely don't appear to be applicable for solving the general quintic, since he does not involve transcendental functions. $\endgroup$
    – ZKG
    Commented Aug 7, 2019 at 19:15
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    $\begingroup$ It appears that the question "Is it true that Descartes left solving the quintic, and higher degree polynomials as an exercise to his readers?" gets a negative answer. Brown's assertion is a half truth or perhaps just bad rhetoric. $\endgroup$
    – sand1
    Commented Aug 7, 2019 at 19:45

It is easy to find fault with Descartes and Leibniz spent his life doing it (see Belaval Y., Lz critique de Desc., P.1960). Descartes knew that some problems of higher degrees are reductible and erroneously believed that it is the general case. The question here however concerns a paraphrase without reference and asks for a good match. such as e.g. La Geometrie p.192. An other one could be p.43.

Mais parce que i’espere que d’orenavant ceux qui auront l’adresse de se servir du calcul Geometrique icy proposé, ne trouueront pas assés de quoy s’arester touchant les problesmes plans, ou solides ; je croy qu’il est à propos que je les invite à d’autres recherches, où ils ne manqueront iamais d’exercice.

= I hope that hereafter those who are clever enough to make use of the geometric methods herein suggested will find no great difficulty in applying them to plane or solid problems. I therefore think it proper to suggest to such a more extended line of investigation which will furnish abundant opportunities for practice.

The text here suggests that more simple case have been rendered trivial and invites the readers to exercice themselves with more complicated ones, promising that they will never lack "opportunity for practice". This might also be a reference for the distorted paraphrase.


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