Update (December 1, 2023)
Few entries have had as much impact as this one. First of all Professor Paul Monsky of Brandeis University wrote the note he reports in his comments. It is a delightful read. He demonstrates the classical results on the subject using simple tools (the most elementary part of algebraic field theory) but with an elegance that I would recommend to all students of mathematics. It blew me away. It is accessible to anyone at this link The sum of two cubes problem — an approach that’s classroom friendly. Paul Monsky, like many mathematicians is very consistent with his ideas. In the 1970s he stopped paying U.S. federal income tax in protest against military spending, for which he was prosecuted. Long live brave (or consistent) mathematicians!
I just don’t think I should be forced to pay for instruments of destruction. I have no illusions about the nature of Russian imperialism or American imperialism, but we already have the ability to destroy the world two times over. (Monsky, March 10, 1980).
This eventually turned into a positive experience: “the judge was sympathetic and allowed me to fulfill my sentence by continuing working shifts from time to time at a shelter for homeless women.”
Monsky has proved many theorems but there is one that anyone can understand, we cannot divide a square into an odd number of triangles of equal area.
The paper by Professor Hongbo Yin that was the initial reason for the entry was finally published in the Transactions of the American Mathematical Society in 2022. But the result he publishes is improved over the version in arXiv that I handled to write the entry. Now in November 2023 he must have heard about my entry and writes to me to comment on this. His polynomial, which I did not write because it was too long, is now reduced to half a line. His theorem is worth writing in full:
Theorem.Let \(p\equiv8\bmod9\) be a prime number such that the equation $$x^9-24x^6+3x^3+1-9(\root 3\of 3-1) x^2(x^3+1)^2=0$$ has no solutions in \(\mathbb F_p\), then at least one of the two numbers \(p\) or \(p^2\) is the sum of two cubes}.
The sequence of primes satisfying this condition start with
17, 53, 71, 107, 179, 233, 251, 269, 359, 431, 503, 521, 557, 593, 701, 719, 773, 809, 827, 863, 881, 953, 971, 1097, 1151, 1187, 1223, 1259, 1277, 1439, 1511, 1619, 1637, 1709, 1871, 1889, 1997, 2087, 2141, 2213, 2267, 2357, 2393, 2447, 2591, 2609, 2699, 2753, 2789, 2843, 2861, 2879, 2897, 2969, 3041, 3167, 3203, 3221, 3257, 3329, 3347, . . .
and still is not in OEIS.
As this entry is published on \(\pi\) Day, we will celebrate it with a composition by mathematician, composer and pianist Noam Elkies entitled Allegro a la J. S. Bach and C. M. von Weber; it encodes the first 244 decimal digits of \(\pi\): every fourth note of the right hand gives a successive digit of \(\pi\), with an appropriate encoding. To find out what this has to do with the substance of our entry, you’ll have to read on.
The Euclid Porisms
Our story begins with a lost book. The Euclid’s Porisms. We know of their existence because Diophantus cites three of the Porisms in his book on arithmetic and Pappus of Alexandria cites others of a geometrical nature in which the first steps of projective geometry can be glimpsed. Diophantus and Pappus lived about the third and fourth century AD. For this reason the Porisms must not have been lost when in 48 Julius Caesar burnt his own ships in the port of Alexandria and the fire destroyed part of the Library. We do not know when it was definitely lost, possibly Hypatia could have seen it. In any case it was lost because of religious fanaticism. Either it was the Christians under Cyril in 415 or the Moors under Caliph Omar in 642 who decided to burn all the manuscripts. If they were in accordance with the Koran they were not necessary and if they were not it was better to destroy them.
We are interested in one of Euclid’s arithmetical Porisms quoted by Diophantus.
Porism If a number \(n\) is the difference of two cubes, then it is the sum of two cubes.
I recall that for Diophantus numbers are positive rational numbers. Diophantus did not have today’s notation for equations. So what he does is to solve the problem in a particular case leaving the reader to see that his method is either general or applies in certain cases. As an example I will use Diophantus’ methods to explain how he would have done it. (This is just my reconstruction, since Diophantus only quotes Euclid).
Let the given number \(n=7 =2^3-1^3\), we are asked to write \(7\) as the sum of two cubes. The first of the cubes will be of side somewhat smaller than \(2\), let’s say \(2-x\), we will take the second cube with side \(mx-1\). So we find that \(7=(2-x)^3+(mx-1)^3\). Developing \begin{align*}(2-x)^3+(mx-1)^3 &= 8-12x+6x^2-x^3\\&-1+3mx-3m^2x^2+m^3x^3\\ \end{align*} We choose \(m\) so that the coefficient of \(x\) cancels out, that is \(m=4\). We will have that $$(2-x)^3+(4x-1)^3=7-42x^2+63x^3.$$ Since we want the sum of these two cubes to be \(7\), we must now take \(x\) so that \(63x^3-42x^2=0\), that is \(x=\frac{42}{63}\), and we obtain the equality $$7=\Bigl(\frac{84}{63}\Bigr)^3+\Bigl(\frac{105}{63}\Bigr)^3.$$ Simplifying $$7=\Bigl(\frac{4}{3}\Bigr)^3+\Bigl(\frac{5}{3}\Bigr)^3.$$
The Porism is not exactly a theorem, nor a corollary, it is a procedure to obtain a result from some data. I leave it to the reader to use the method in general for \(n=a^3-b^3\) where \(a\) and \(b\) are positive rationals and to find under what condition we obtain \(n=c^3+d^3\) with \(c\) and \(d\) positive rationals.
Two kinds of numbers
We have seen that \(7\) is a sum of two cubes, not every number has this property. We cannot write \(1\) as a sum of two cubes. If we find two rational numbers with \(1=a^3+b^3\), we could write the two fractions with a common denominator \(a=\frac{x}{z}\), \(b=\frac{y}{z}\), and then \(z^3=x^3+y^3\). But Wiles’ theorem tells us that this is not possible with non-zero numbers. That is to say \(1\) is not the sum of two non-zero cubes. The sequence A020898 is the sequence of the integers \(n\), free of cubes, such that the equation \(x^3+y^3=nz^3\) has non-trivial solutions.
A quarrelsome mathematician?
In 1841, newly recruited by the University of Virginia, Sylvester was eager to continue his mathematical research. He was then 27 years old and had a reputation as one of the best English mathematicians. Surprisingly, after 5 months, he rushed back to London without even picking up his books in Virginia. What made him give up his project? We could say that he stumbled over a pupil. The pupil had repeatedly annoyed Sylvester during his classes. Sylvester had complained to the faculty, who in turn had reprimanded the student. On the last occasion when Sylvester reprimanded him, the pupil felt his honour had been wounded, and challenged him to a duel. When they met, Sylvester carried a sword cane and the pupil a heavy baton. The pupil stood in front of him and asked him to apologise, with the truncheon he knocked off his hat and then struck him on the head. Sylvester drew his sword and lunged at the student. He hit him right in the heart, the student shouted, “I’m dead! He killed me!” On the advice of one of those present Sylvester ran out…, a doctor came and found the pupil livid and bathed in sweat, opening his shirt he saw that there was only a scratch. Sylvester had touched bone. The event put Sylvester out of mathematics for 12 years, during which time he worked as an actuary in London and studied law. He then met Cayley, another lawyer by profession but a dedicated mathematician.
The object of law was to say a thing in the greatest number of words, of mathematics to say it in the fewest.
Arthur Cayley
The problem we are dealing with today was considered by Sylvester in his second mathematical period, after the clash with the student. In a general form, we could put the problem as follows:
Sylvester’s problem. Which natural numbers \(n\) can be decomposed as a sum of two non-zero rational cubes?
If \(n=a^3+b^3\), we can write \(a\) and \(b\) with a common denominator \(a=x/z\), \(b=x/z\) and we arrive at the equivalent problem:
Problem. For what values of \(n\) does the diophantine equation \(x^3+y^3=nz^3\) have non-trivial solutions? That is, other than \((1,-1,0)\) and its multiples.
The problem is prior to Sylvester. Legendre had proved that in the case \(n=1\), \(3\), \(4\) and \(5\) the equation \(n=a^3+b^3\) has no solutions, and he thought that the same would be true in the case of \(n=6\). But alas $$6= \Bigl(\frac{37}{21}\Bigr)^3+\Bigl(\frac{17}{21}\Bigr)^3.$$ Sylvester manages to prove for many numbers \(n\) that the equation has no solutions.
Sylvester posed the problem of proving some of his results in the journal Nouvelles annales de mathématiques, the journal of the Candidates to the École Polytechnique Normale. Mathematician Lucas published a solution in the journal that greatly simplifies Sylvester’s arguments and gives a theorem that I find very attractive:
Theorem. (Lucas) The equation \(x^3+y^3=nz^3\) has non-trivial solutions if and only if there exist integers such that \(n=uv(u+v)/w^3\).
Proof. Indeed if \(n w^3=uv(u+v)\), it is easy to check that $$(u^3-v^3+6u^2v+3uv^2)^3+(v^3-u^3+6v^2u+3vu^2)^3=nw^3(u^2+uv+v^2)^3.$$ On the other hand if \(x^3+y^3=nz^3\), we take \(u=x^3\), \(v=y^3\) and then $$uv(u+v)=x^3 y^3(x^3+y^3)=x^3 y^3 n z^3=n(x y z)^3.$$ and it suffices to take \(w=xyz\). qed.
In general this is very difficult to apply. But in the case of \(n=6\) we have \(6=2\times 3=1\times2\times(1+2)\) and we obtain the aforementioned solution that Legendre did not find.
The results of this first period of the problem can be summarised in the theorem
Theorem. (Pepin, Lucas, Sylvester) If \(p\equiv 2\) or \(p\equiv 5 \bmod9\) is an odd prime then neither \(p\) nor \(p^2\) can be written as a sum of two cubes.
The modern theory of Sylvester’s problem
We want to know if there are rational points on the curve \(x^3+y^3=n\). We are in luck, it is an elliptic curve \(E_n\). The rational points of an elliptic curve form an abelian group. If \(P\) is a rational point on \(E_n\) we can construct other \(2P=P+P\), \(3P\), and so on. This is what Euclid already intuited, since he could construct a rational point \(n=A^3+B^3\) from another \(n=a^3+(-b)^3\).
If the multiples of \(P\) form a finite subgroup, the point \(P\) is called a torsion point. The curves \(E_n\) have no torsion except for \(n=c^3\) or \(n=2c^3\), we can forget about these cases and then our question is related to the rank of the curve. If the range is \(0\) the curve has no rational points. If the range is \(\ge1\) it will have infinite rational points.
Selmer is the one who opened this point of view, and he succeeds in proving $$\text{Rank of $E_p$} \le\begin{cases}0, &\text{if $p\equiv2, 5\pmod 9$},\\ 1, &\text{if $p\equiv4,7,8\pmod9$},\\2, &\text{if $p\equiv1 \pmod9$}.\end{cases}$$ One of the millennium problems, each endowed with a million euro prize (and what’s that, when some get 65 for a much less complicated task), has to do with this. The Birch and Swinnerton-Dyer conjecture.
Conjecture (of Birch and Swinnerton-Dyer). Let \(E\) be an elliptic curve and \(L(E,s)\) be the associated \(L\) function. The rank of the curve \(E\) is equal to the order at the point \(s=1\) of the function \(L(E,s)\).
Therefore the conjecture predicts that if \(L(E,1)\ne0\), then the curve has no rational non-torsion points. On the other hand if \(L(E,1)=0\) then the curve has infinitely many rational points. As we can see Selmer’s calculation already proves that for \(p\) odd prime \(\equiv 2,5\bmod 9\) there will be no rational points in \(E_p\).
When \(p\equiv 4,7,8\bmod9\), it is known that the sign of the functional equation of \(L(E_p,s)\) is \(-1\). This implies that \(L(E_p,1)=0\) in these three cases and if the Birch and Swinnerton-Dyer conjecture is true there will be infinitely many rational solutions of the equation \(x^3+y^3=p\).
The story breaks down here into several cases. Many of them depending on the techniques introduced by an amateur mathematician whose merits were recognised only after his death.
Heegner’s points
Heegner (1893-1965) worked mainly as an engineer, obtaining several patents in the field of radio. They were used by Telefunken. As a mathematician he worked in isolation, although he had an adequate training at the University of Berlin. Hecke was referee of one of his papers, claiming that Heegner understood the subject well but that his exposition was unfortunate, using his own, in many cases superfluous, terminology.
His work (1952) was the result of a new technique for obtaining rational points on curves. With it he manages to solve an important problem proposed by Gauss: on which imaginary quadratic fields does the fundamental theorem of arithmetic hold? Heegner was able to prove that precisely on \(\mathbb Q(\sqrt{n})\) for \(n=-3\), \(-4\),\(-7\),\(-8\),\(-11\), \(-19\), \(-43\), \(-67\) and \(-163\).
In his proof Heegner used a book by H. Weber, Lehrbuch der Algebra, which contained some errors. This together with the obscurity of his exposition made Heegner’s work unacceptable to the mathematical community. Baker (1966) and Stark (1967) solved the problem again. It was at this time that Birch, Deuring, Stark made it clear that in fact Heegner’s proof was correct, if only with a small gap that was easy to fill.
Heegner’s idea is roughly the following: Any elliptic curve admits a parametrization by modular forms. The image of the imaginary quadratic numbers is algebraic. The sum on the elliptic curve of all conjugates is a candidate, since it will be on the curve and invariant for the Galois group. These are the so-called Heegner points.
Recent results
The first result is due to N. Elkies who announced the proof that in the case \(p\equiv 4,7\bmod9\) the rank of the curve \(E_p\) is \(1\), by a modification of Heegner’s method for finding points. The details of Elkies’ proof have not been published. Concerning this same case S. Dasgupta and J. Voight (2006) prove, (also using Heegner’s techniques but in a different way than Elkies), that the rank of \(E_p\) and that of \(E_{p^2}\) is 1 whenever \(3\) is not a cube modulo \(p\).
Elkies is not only a great mathematician but also a composer and pianist. Sometimes he gives a mathematical vision of music: one of his compositions is precisely the work Allegro a la J. S. Bach and C. M. von Weber which encodes the first 244 decimal digits of \(\pi\) and of which we included some bars at the beginning of this entry.
The case \(p\equiv1\bmod 9\) has been treated by Zagier and Villegas (1995), in this case there are two possibilities: either the rank of \(E_p\) is \(0\) or it is \(2\) (giving rise to equations \(x^3+y^3=p\) without or with solutions respectively).
The theorem the proved is
Theorem (Zagier, Villegas). We define a sequence of numbers \(c_0=1\), \(c_1=2\), \(c_3=-152\), … through \(c_n=s_n(0)\) where \(s_0(x)=1\), \(s_1(x)=3x^2\) and $$s_{n+1}(x)=(1-8x^3)s’_n(x)+(16n+3)x^2 s_n(x)-4n(2n-1)x s_{n-1}(x),\quad n\ge1.$$ If \(p=9k+1\) is prime, then \(L(E_p,1)=0\) if and only if \(p\) divides \(c_k\).
Fernando Villegas is an Argentinian mathematician who has worked in the United States since his youth. His mathematics is always very attractive and surprising. We have already talked about Don Zagier in a previous post.
Villegas and Zagier’s results use techniques different from Heegner’s points, associated rather with the \(L\) function of the elliptic curve.
After this the difficult case is the \(p\equiv8 \bmod9\) case.
On 31 December 2019 Hongbo Yin uploaded to arXiv his paper in which he proves the following result:
Theorem (Hongbo Yin). Let \(p\equiv8\bmod9\) be a prime. There exists an explicit polynomial \(D(x)\) of degree 27 such that the following holds: If \(3\mid \frac{p+1}{9}\) and \(D(x)\) does not cancel modulo \(p\) or else if \(3\) does not divide \(\frac{p+1}{9}\) and \(D(x)\) cancels modulo \(p\), then at least one of the two \(p\) or \(p^2\) can be written as a sum of two rational cubes.
I do not write the polynomial that spans three lines in Hongbo Yin’s paper.
Therefore the difficult case is not satisfactorily solved so far. It must be said that Elkies when announcing his work added a table with the solutions of the equation in the cases \(p\equiv 4, 7\bmod9\) provided by his method. Elkies also applied his method to the case \(p\equiv 8\bmod 9\). Although he was not able to prove it in general, his method gives him the solution of \(p=x^3+y^3\) for all the values of \(p\equiv 8\) that he was able to test on his computer.
Learn more
Sylvester is certainly an interesting character, there is a book of his correspondence which shows that he was connected with the most important personalities of the time.
K. H. Parshall, James Joseph Sylvester. Life and Work in Letters, Oxford Univ. Press, 1998.
But the story of his encounter with the wayward pupil is told in
L. S. Feuer, Sylvester in Virginia, The Mathematical Intelligencer 9 (1987), 13-19.
Sylvester’s works are difficult to read, but I highly recommend the paper by E. Lucas, Sur l’Analyse Indéterminée du Troisième Degré.-Demonstration de Plusieurs Théorèmes de M. Sylvester, American J. Math., 2, (1879) 178-185
About Heegner we found the PDF of a lecture by N. Schappacher on his life Kurt Heegner.
A great deal of information on the problem can be obtained from Noam Elkies’ pages: Tables of solutions… and the table of experimental solutions for the case \(p\equiv8\bmod9\). Information about the musical activities of Noam Elkies But the one we have taken about the number \(\pi\) is taken from the website Allegro a la J.S. Bach and C.M. von Weber.
The original works we have consulted on the subject are not easy reading. The most accessible is the book by Zagier
J. H. Bruinier, G. van der Geer, G. Harder y D. Zagier, The 1-2-3 of Modular forms, Springer, 2008, p.~97-99.
Selmer’s paper inaugurates the new period of the E. S. Selmer problem., The Diophantine Equation \(ax^3+by^3+cz^3=0\), Acta Math. 85 (1951) 203-362, and E. S. Selmer, The Diophantine Equation \(ax^3+by^3+cz^3=0\). Completion of the tables, Acta Math. 92 (1954) 191-197.
F. Villegas and D. Zagier, Which primes are sums of two cubes?, Number theory (Halifax, NS, 1994), 295-306, CMS Conf. Proc., 15, Amer. Math. Soc., Providence, RI, 1995.
S. Dasgupta and J. Voight, Sylvester Problem and Mock Heegner points, Proc. Amer. Math. Soc. 146 (2018) 3257-3273.
Hogbo Yin, On the $8$ case of Sylvester Conjecture, arXiv:1912.13338.
Thank you for this delightful article. I’ve recently found nice simplifications of Sylvester’s non-existence proof, using nothing beyond simple facts about the Eisenstein integers. I’m writing them up for a note that should be accessible to undergrad math majors with an interest in number theory and algebra (but I eschew the language of ideals and elliptic curves). I’ve put references to your article in my note to give readers some historical perspective. I’ve also shown that certain irreducible Eisenstein integers aren’t (x^3) + (y^3) with x and y in Q[omega]. And given a little empirical evidence for one existence conjecture. I’ll send you the note soon if you’re interested.
A question for the author—A couple of nights ago I left a comment praising his article, telling him that I’d be including references to it in some notes I was writing up giving extremely simple proofs
of classical results of Euler, Legendre, Kummer and Sylvester involving the sum of two cubes problem, and giving a few new results with the field Q replaced by the 3-cyclotomic field. I asked the author if he’d like to see these notes when I completed them in the near future. But I got no response, and my comment has since disappeared! I assure him that it’s not junk, and again ask if he’d like to see the notes.
Yes, I would like to see, sorry I do not see the comments with a periodicity.
Greetings, Professor de Reyna
I finished the notes(35 handwritten pages), and will send you a copy in the mail if you provide an address. It ends with a flourish–a pretty result that Sylvester missed, and I think it will entertain you.
Here are the new results:
1. If m in Z[omega] is irreducible of norm that’s 4 or 7 mod 9, then no associate of m or m^2 (apart from the obvious exceptions of 2 and -2) is (x^3) + (y^3) with x and y in K = Q[omega].
2. Let p in Z be a prime that’s 1 mod 9. Let m be either of the two irreducible factors of p in Z[omega] that are 1 mod 3. Then when m isn’t 1 mod 9, neither m nor m^2 is (x^3) + (y^3) with x and y in K.
So for example neither -(5 + 3*omega) nor 4 – 3*omega is a two cube sum. When m is 1 mod 9, it seems plausible that m and m^2 are two cube sums.But the curves have even K- ranks, so I doubt that the Heegner approach will produce solutions. What’s nice is that all the old results as well as the new have the same basic simple proof. I’m still writing things up—what address should I send them to when I’m done? Thanks, Paul.
Diophantus does cite the Porisms three times in his Book V, but he does not cite Euclid. In fact, most historians believe that Diophantus is talking about a work of his own.
Let me also mention that in Euclid’s books on number theory, arithmos always means “natural number”; for this reason alone I think that this result cannot be credited to Euclid.
First some facts:
(A) Diophantus phrase is We have it in the Porisms that \dots and yes he uses it three times in his book of Arithmetic.
(B) There is a (lost) book of Euclid with the title Porisms. We know by Pappus Porisms of Euclid consists of three books and 171 propositions. Pappus speaks about some of these porisms and they are of geometrical character.
(C) In modern times many authors speculated that the porisms were simply part of the Arithmetica, others assumed the Porisms in Diophantus alluded to a lost book of Diophantus. But there is no old source speaking about a book by Diophantus with that title.
Fact (C) is difficult to explain. How a book that is quoted by Diophantus simply by his title was not cited by anyone. So Occam’s razor principle leads us to suggest that it is the book of Euclid.
Diophantus cite the Porisms as a well known book. He does not have to talk about Euclid if the source is clear. But if he refers to another book with the same title (written by himself or any other not Euclid) he should have specified.
The question of language about the meaning of arithmos in Euclid requires a comment. Did Euclid know that $latex \sqrt{2}$ is not rational? Of course he uses other words for this. Diophantus did know about rational numbers and it is quite possible that he is reformulating a result of Euclid.
Euclid expose the theory of proportions of Eudoxus, precisely because not all proportions are rational.
Euclid’s Porisms is a geometrical book, but this is also true for the Elements. The Porisms could contain arithmetic results. More when this arithmetical result read the difference of two cubes can be transformed into the sum of two cubes. Equivalently: Given two cubes (by his sides) construct other two (by his sides) so that the difference of the two given cubes is equivalent to the sum of the two constructed ones.
In fact the sides of the cubes in the porisms are constructible with rule and compass and the construction shows to Diophantus that the sides constructed are rational if the data sides are. When $latex a^3\ge 2b^3 $ the construction is $$a^3-b^3=a^3 \Bigl(\frac{a^3-2b^3}{a^3+b^3}\Bigr)^3+b^3\Bigl(\frac{2a^3-b^3}{a^3+b^3}\Bigr)^3.$$
Dear Professor de Reyna, thank you for your so nice blog explaining Sylvester’s problem and citing my result. I just like to inform you that the condition in my result has been simplied dramastically in my second (also published) version. Please have a look if you have interest.