Egyptian Mathematics

In the fourth month of the flood season in the 33rd year of the reign of Pharaoh Apophis I (16th century BC) the scribe Ahmes was copying on parchment a collection of mathematical problems. The collection dated from the time of Pharaoh Amenemhat III (19th century BC). That it has been preserved to this day is a true miracle. It shows us a mathematics that is not as advanced as that of the Plimpton 322 cuneiform tablet, which may be from the same period as Ahmose, but a mathematics that is decidedly interesting and that may leave us puzzled at first. You will see why I say so.

When we want to explain fractions to a small child, it is inevitable to talk about how we divide up a cake. If there are 5 children, we divide the cake into 5 equal parts and each child receives a fifth, \(\frac15\) of the cake. This way we always end up with unit fractions. The Egyptians never wanted to go beyond this stage. Their numbers looked like $$\frac14+\frac18+\frac{1}{10} +\frac{1}{30}+\frac{1}{45}.$$ One of the problems posed in the papyrus is to find the difference of \(1-\frac{1}{3}\) and the previous number. The solution they get is $$\Bigl(1-\frac{1}{3}\Bigr)-\Bigl(\frac14+\frac18+\frac{1}{10} +\frac{1}{30}+\frac{1}{45}\Bigr)=\frac{1}{9}+\frac{1}{40}.$$ That is to say, they find that $$\frac13+\frac14+\frac18+\frac{1}{9}+\frac{1}{10} +\frac{1}{30}+\frac{1}{40}+\frac{1}{45}=1.$$ It is certainly more interesting mathematics than that of the Romans.

Writing rational numbers in this way is a rather inexplicable quirk. They only handled unit fractions, with one exception: \(\frac23\) was allowed. This tells us that they knew other fractions and didn’t want to use them. Any child would think of taking \(\frac37\) from the cake and leaving the other 4 children a piece of \(\frac17\) each, because that’s what their birthday is for. No, the Egyptians considered \(3\cdot\frac17\) as a problem: Find a number that multiplied by 7 gives 3. An Egyptian would proceed by approximations like this \begin{align*} 7\times \frac13&=2+\frac13 && \text{it is not still 3 but it’s close,}\\ 7\times \frac{1}{14}&= \frac12 && \text{still a little way to 3,}\\ 7\times \frac{1}{42}&=\frac16 &&\text{got it!} \end{align*} Then $$7\times\Bigl(\frac13+\frac{1}{14}+\frac{1}{42}\Bigr)=3.$$ So $$ 3\times\frac{1}{7}=\frac13+\frac{1}{14}+\frac{1}{42}.$$ Are you getting the hang of Egyptian mathematics? The reason why they prefer the number \(\frac13+\frac{1}{14}+\frac{1}{42}\) to our fraction \(\frac{3}{7}\) remains totally elusive, if we do not think of Mathematics as what it was in its beginnings. The supreme game, in this case with numbers, in others with geometric figures. 

What do we find in the papyrus? Not rules, but examples of how to operate with numbers. Numbers are sums of unit fractions written in decreasing order, so \(\frac13+\frac{1}{14}+\frac{1}{42}\) is a number, but \(\frac17+\frac17+\frac{1}{42}\) is not (repeated fractions are not allowed). As I said, they allowed \(\frac23\), but I will overlook this discrepancy. 

Challenge: Prove that any problem \(m\times\frac1n\) has an acceptable solution for an Egyptian. In other words any positive rational number \(\frac{m}{n}\) is the sum of a finite number of distinct unit fractions. 

The first problem that appears in the papyrus is that of doubling a unit fraction, calculating \(2\times \frac{1}{n}\). Its methods would be of the following type: (1) Find a number \(a\) with many divisors and such that \(n<2a\). (2) Let \(c=2a-n>0\). We decompose \(c\) as a sum of different divisors of \(a\). Let be for exampe \(c=d_1+d_2+\cdots+d_k\) $$\frac{c}{a}=\frac{1}{a/d_1}+\frac{1}{a/d_2}+\cdots+\frac{1}{a/d_k},$$ then $$\frac{2a-n}{na}=\frac{c}{an}=\frac{1}{n(a/d_1)}+\frac{1}{n(a/d_2)}+\cdots+\frac{1}{n(a/d_k)}.$$ That is, $$2\times\frac{1}{n}=\frac{1}{a}+\frac{1}{n(a/d_1)}+\frac{1}{n(a/d_2)}+\cdots+\frac{1}{n(a/d_k)}$$ is the wanted number. To double \(\frac{1}{17}\) we take \(a=10\) and then \(20-17=3\). We write \(3=2+1\), then \(\frac{3}{10}=\frac15+\frac{1}{10}\) and the solution is $$2\times\frac{1}{17}=\frac{1}{10}+\frac{1}{85}+\frac{1}{170}.$$ We notice that the trick can be generalised to other fractions.

We leave the Rhind papyrus not without mentioning that other problems are concerned with calculating the volume of a cylindrical silo, which leads to a good approximation of the area of the circle. Anyone who is more interested in Egyptian mathematics can look in the last section, where we give some very good references. 

Egyptian fractions today

There are plenty of publications, in serious and other minor publications, on issues related to the Egyptian fractions. But today we want to focus on the following problem:

Erdös-Straus problem. To prove that for any natural number \(n\ge2\) there exist three natural numbers \(x\), \(y\), \(z\) such that $$\frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$ In this case we accept that some of the three numbers can be equal. The tricky issue here is that we want three fractions and no more. Two fractions will also do since \(\frac{1}{x}=\frac{1}{2x}+\frac{1}{2x}\) and we can turn a solution with two fractions into one with three.

If I have a solution for \(n\), I have a solution for any multiple of \(n\) since \(\frac{4}{mn}=\frac{1}{mx}+\frac{1}{my}+\frac{1}{mz}\) solves the problem for \(mn\) if we know how to solve it for \(n\). Therefore the problem is only in the case where \(n\) is prime. Since in addition \(2=\frac{4}{2}=1+\frac12+\frac12\), from now on we can assume that \(n\) is an odd prime number.

In general there are many solutions. For example, for \(n=7\) the possible solutions are $$\frac{4}{7}=\frac12+\frac{1}{15}+\frac{1}{210}=\frac12+\frac{1}{16}+\frac{1}{112}$$ $$=\frac12+\frac{1}{18}+\frac{1}{63}=\frac12+\frac{1}{21}+\frac{1}{42}$$ $$=\frac12+\frac{1}{28}+\frac{1}{28}=\frac13+\frac{1}{6}+\frac{1}{14}$$ $$=\frac14+\frac{1}{4}+\frac{1}{14}.$$ This is a problem reminiscent of Egyptian mathematics but it is actually a Hungarian problem. We have already discussed matemática húngara in a previous post. The problem was posed in writing in 1950 in two papers, one by Erdös in Hungarian and the other by Obláth in a non-digitised journal. I have not been able to consult either of them, but Christian Elsholtz confirms that Obláth’s paper already refers to the problem as the Erdös-Straus problem.

Each solution can be generalised.  For example, the last one we gave for \(n=7\),  \(\frac{4}{7}=\frac14+\frac{1}{4}+\frac{1}{14}\) is a particular case of $$\frac{4}{4k+3}=\frac{1}{2k+2}+\frac{1}{2k+2}+\frac{1}{(k+1)(4k+3)}.$$ Roughly half of the primes give remainder \(3\) when divided by \(4\) and fall within this scheme. Could we find an analogous solution for \(\frac{1}{4k+1}\)? We would have solved the problem, since every odd prime gives remainder \(1\) or \(3\) when divided by \(4\). If we found three polynomials \(P_1(n)\), \(P_2(n)\) y \(P_3(n)\) such that $$\frac{4}{4k+1}=\frac{1}{P_1(k)}+\frac{1}{P_2(k)}+\frac{1}{P_3(k)}$$ and taking integer values for \(k\) integer we would have solved our problem. This hope vanishes with a Schinzel theorem:

Theorem. Let \(a\) and \(b\) be integers, \(a>0\), prime to each other (\(\textrm{mcd}(a,b)=1\)) and such that \(b\) is a quadratic remainder modulo \(a\). Then there do not exist polynomials \(P_j(x)\in\mathbb Z[x]\) with positive leading coefficients that satisfy $$\frac{4}{an+b}=\frac{1}{P_1(n)}+\frac{1}{P_2(n)}+\frac{1}{P_3(n)}.$$

However, this is not the end of the matter. Many mathematicians have attacked the problem over the years. For example, Mordell’s classic book summarises the situation in 1969 as follows:

The conjecture is true except at most in cases where \(n\) is congruent to \(1^2\), \(11^2\), \(13^2\), \(17^2\), \(19^2\) o \(23^2\) modulo \(840\).

We will never be able to solve the problem with only one modulo because of Schinzel’s theorem.

Congruences solvable by polynomials

We have seen that for the primes \(n\equiv3\bmod4\) the Erdös-Straus problem is solved by polynomials but for the primes \(n\equiv1\bmod4\) an analogous solution is not possible. The general solution to the question of which congruences admit a solution by polynomials is due to Christian Elsholtz and Terence Tao. Let us give a congruence \(n\equiv r \bmod N\) containing infinitely many primes, i.e., thanks to a Dirichlet theorem such that \(r\) and \(N\) be prime to each other. Are there polynomials \(P_1(n)\), \(P_2(n)\), \(P_3(n)\) such that they take integer values for \(n\equiv r\bmod N\) and furthermore $$\frac{4}{n}=\frac{1}{P_1(n)}+\frac{1}{P_2(n)}+\frac{1}{P_2(n)}?$$ The complete answer is given by Elsholtz and Tao by means of a theorem that states that \(n \equiv r \bmod N\) is solvable by polynomials if and only if it falls into one of 7 possibilities. We are not going to describe all of them, as an example we will give only two:

(1) If the congruence can be written in the form \(n\equiv -f\bmod{4ad}\) with \(f\mid 4a^2d+1\). In this case the solution is $$\frac{4}{n}=\frac{1}{a(\frac{n+f}{4ad}\frac{4a^2d+1}{f}-a)dn}+\frac{1}{a(\frac{n+f}{4ad})d}+\frac{1}{(\frac{n+f}{4ad}\frac{4a^2d+1}{f}-a)(\frac{n+f}{4ad})d}.$$ No great mystery here, the above equation is an identity and the conditions allow us to see that the denominators are natural numbers.

(2) There are three numbers \(a\), \(c\) and \(f\) with \(4ac\) prime with \(f\) such that the congruence is equivalent to the two conditions \(n\equiv-f\bmod{4ac}\) y \(f\mid an+c\). In this case $$\frac{4}{n}=\frac{1}{a(\frac{an+c}{f})(\frac{n+f}{4ac})n}+\frac{1}{ac(\frac{n+f}{4ac})}+\frac{1}{(\frac{an+c}{f})(\frac{n+f}{4ac})c}.$$ As before, this is an identity and the conditions guarantee that the denominators are integers.

The difficult part is to see that the only resolvable congruences are the 7 indicated in the theorem, of which these are two examples. The main goal of Elsholtz and Tao is not this theorem, but neither do they solve the conjecture. What they do achieve is to establish the average number of solutions \((x,y,z)\) of the Erdös-Straus equation for each prime \(n\). The conjecture says that there is at least one solution for each prime, but in general there are many. If \(f(n)\) denotes this number of solutions they prove that $$N\log^2 N \le \sum_{p\le N}f(p)\le N\log^2N\log\log N.$$ We can summarise the above by saying that on average a prime \(p\) admits \(\log^3p\) solutions. But this is a typical case of the paradox of the average: On average each prime \(p\) admits \(\log^3p\) solutions, but a typical prime has many less. (It is not the same to talk about the average wealth of the Spaniards than about the wealth of the average Spaniard).  

In the proof they separate the solutions into two types. Type I solutions verify that \(n\) divides \(x\) but not \(y\) or \(z\). In type II solutions \(n\) divides \(y\) and \(z\) but not \(x\). For \(n\) prime there are only these two types of solutions. In each solution there are hidden variables that are very useful. For example, solutions of type I can always be written in the form $$\frac{4}{n}=\frac{1}{abdn}+\frac{1}{acd}+\frac{1}{bcd}.$$ It is enough to take \(d=\textrm{mcd}(x,y,z)\) and having \(x’=x/nd\), \(y’=y/d\), \(z’=z/d\), take \(a=\textrm{mcd}(x’,y’)\), \(b=\textrm{mcd}(x’,z’)\), \(c=\textrm{mcd}(y’,z’)\). Finally we define \(f=4acd-n\), and it turns out that \(a+b\) is then divisible by \(c\) and we define \(e=(a+b)/c\). 

For example, for $$\frac{4}{7}=\frac{1}{63}+\frac{1}{18}+\frac12,\qquad (a,b,c,d,e,f)=(9,1,2,1,5,65).$$ These numbers verify many relations: $$4abd=ne+1,\quad ce=a+b,\quad4abcd=na+nb+c,\quad 4acde=ne+4a^2d+1,\quad 4bcde=ne+4b^2d+1,$$ $$4acd=n+f,\quad ef=4a^2d+1, \quad f=na+c,\quad n^2+4c^2d=f(4cd-n).$$ The reciprocal is true, numbers satisfying these relations give us a type I solution. This then allows us to count the solutions by counting these parameters. To do this we end up estimating arithmetic sums. For example one of the sums that need to be estimated is $$\sum_{a\le A, b\le B} \tau(kab^2+1)\ll AB\log(A+B)\log(1+k)$$ with certain conditions on \(k\) and where \(\tau(n)\) denotes the number of divisors of the number \(n\). 

Erdös raised many problems, he liked to talk about mathematics with anyone interested. On many occasions with children. He is pictured here with Terence Tao, who was a child prodigy. Over the years Tao has worked on many of Erdös’ problems.

Christian Elsholtz is a professor at the Technical University of Graz in Austria, interested in number theory, combinatorics and analytics. He has numerous papers on sums of unit fractions.

The work by Salez

We now focus on a later article by Serge E. Salez (The Erdös Straus conjecture. New modular equations and checking up to \(N=10^{17}\)). It is an interesting paper proving that the Erdös-Straus conjecture is valid for all \(N\le 10^{17}\). The paper is interesting because one cannot verify all these cases by finding a solution for each case. It is then necessary to find a method that reduces the question to a practical number of checks.

We have already seen that in 1956 it was known that the conjecture is true except at most in cases where \(n\) is congruent to \(1^2\), \(11^2\), \(13^2\), \(17^2\), \(19^2\) or \(23^2\) modulo \(840\), then we could consider only \(6\) of each \(840\) numbers. Still this is not enough. Using elementary but clever methods Salez manages to change the number \(840\) to \(G_7=892371480\). Instead of the six numbers \(1^2\), \(11^2\), \(13^2\), \(17^2\), \(19^2\) or \(23^2\), with the modulo \(G_7\) he is left with \(147348\) \(a\) numbers. He now has the task of checking the primes of the form \(n=k\times G_7+a\). For them he determines a set of screenings, i.e. modules \(m\in M\) and for each one a set of remainders \(S_m\) such that if \(n\bmod m\in S_m\) then the Erdös-Straus problem has a solution for \(n\). 

There is a certain mystery about the author of this work. This is the only work he has. In the paper he does not mention any institution. He is not registered as an author in MathSciNet, i.e. he has not published anything so far. The paper is interesting and well written as by a professional. He posted the paper on arXiv in 2014, but it is not published in any journal. I don’t have any information about him.

The BBF conjecture

Manuel Bello Hernández, Manuel Benito, and Emilio Fernández are three Spanish mathematicians from La Rioja. They had proved the Erdös-Straus conjecture for \(n\le 10^{14}\) before Salez did it for \(n\le 10^{17}\). For this they use a polynomial solution which falls into the second scheme of Elsholtz and Tao that we have written above. They put it in the form $$\frac{4}{n}=\frac{1}{(ac-1)(\frac{bc-1}{4})n}+\frac{1}{a(\frac{bc-1}{4})}+\frac{1}{a(ac-1)(\frac{bc-1}{4})},$$ assuming that \(bc\equiv1\bmod4\) and \(n=abc-a-b\). They verify that every prime \(n\equiv1\bmod4\) with \(n\le 10^{14}\) can be written in the form \(n=abc-a-b\) with \(bc\equiv1\bmod4\). Therefore it is natural to consider their conjecture:

BBF Conjeture. Every prime \(n\equiv1\bmod 4\) can be written in the form $$n=abc-a-b.$$ being \(bc\equiv1\bmod4\).

If this conjecture is true then the Erdös-Straus conjecture is true as well. But the reciprocal is not true. There are 7 different ways for the Erdös-Strauss conjecture to be fulfilled and the Bello-Benito-Fernandez (BBF) conjecture is just a very particular one. In fact Elsholtz-Tao remark that the average number of solutions for a prime \(n\) of the BBF conjecture is only \(\log^2n\), while the average number of solutions for the Erdös-Strauss conjecture is \(\log^3n \log\log n\). 

I find the BBF conjecture very interesting. If we want to see if BBF is satisfied for a certain \(n\) we can first test the cases \((a,b)=(1,1)\), \((2,1)\) (remember that \(b\) has to be odd). To do this we check if \(c=\frac{n+a+b}{ab}\) is an integer, and if it is we look to see if \(bc\equiv1\bmod4\). Then we can find a solution. If not, we will have that any solution satisfies \(a\), \(b\ge2\), therefore $$n=abc-a-b\ge ab-a-b=(a-1)(b-1)-1.$$ So we only have to try the pairs \((a,b)\) with \(b\ge3\) odd less than or equal to \(n+2\) and \(a\le (n+1)/(b-1)+1\). If we do not find a solution with these pairs, there is no solution. 

We thus arrive at a more general question than the BBF conjecture: which odd numbers can be put into the form \(n=abc-a-b\) with \(bc\equiv1\bmod4\)?

Trying this process with all the odd numbers, we see that the squares do not satisfy the generalised BBF conjecture, something that Bello, Benito and Fernández had already noticed. Apart from these there are other numbers, it seems that always \(\equiv1\bmod4\) for which the generalised conjecture is false: $$105, 465, 585, 801, 2641, 2929, 2961, 3705, 4785, 4945, 9529, 13041, 13465, 14841,\dots $$ What numbers are these?

I thank Christian Elsholtz for some comments on a previous version of this post.

Learn more

The papers commented in this post are the following:  

Elsholtz, Christian; Tao, Terence, Counting the number of solutions to the Erdős-Straus equation on unit fractions, J. Aust. Math. Soc. 94 (2013), no. 1, 50–105. 

M. Bello Hernández, M. Benito, E. Fernández, On egiptian fractions, arXiv:1010.2035. (2010). 

Serge E. Salez, The Erdős-Straus conjecture New modular equations and checking up to $N=10^{17}$, arXiv:1406.6307 (2014).

On the Rhind Papyrus there is an exceptional book: 

Gay Robins and Charles Shute, The Rhind Mathematical Papyrus an ancient Egytian text, Published by the Trustees of the British Museum by British Museum Publications

There are other publications with better pictures of the Papyrus; also from the British Museum.

The Rhind Mathematical Papyrus British Museum 10057 and 10058. Photgraphic facsimile, Hieroglyphic transcription, transliteration, literal translation, free translation, mathematical commentary and bibliography, in two volumes. 

To anyone who finds it wrong for me to give these inaccessible references I only remind you of what Jesus Christ said:

Ask, and it shall be given you; seek, and ye shall find; knock and the door will be opened to you.  Matthew 7:7.

Update: In October 2022 have been published the survey:

Thomas F. Bloom, Christian Elsholtz, Egyptian Fractions, arXiv:2210.04496.