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Egyptian fractions

The ancient Egyptians were amazing engineers. However, they did not use the mathematical technology that we use today. For instance, they did not write fractions in the form $\frac{p}{q}$. Instead they would write a fraction as a list of numbers $\{x_1,x_2,...,x_n\}$ such that \[\frac{p}{q}=\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}.\] They also worked with the requirement that the list could contain no duplicates! See this website for more information about egyptian fractions in general.

Erdős-Strauss Conjecture

The Erdős-Strauss conjecture is stated as follows:
Note that, if $\frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ for some positive $x,y$, and $z$, then $\frac{4}{kn}=\frac{1}{kx}+\frac{1}{ky}+\frac{1}{kz}$ for any positive integer $k$.
For any number $n$ we can write $n=pk$, where $p$ and $k$ are positive integers and $p$ is some prime. Because of this, a proof of this conjecture only needs to consider when $n$ is a prime number.

Mordel's work

The first resource I found in researching the Erdős-Strauss conjecture was the book Diophantine Equations (1969) by Louis Joel Mordel. In chapter 30 of the book Mordel covers some contemporary results about the conjecture.

I found his proofs to be quite skeletal, and the original proofs he was referencing (which I will also link to when appropriate) were in Italian. Because of this, I will recreate these proofs here in a way that I find suitable. In this writeup, when no confusion arises, I sometimes use $\gcd{a}{b}$ to denote the greatest common divisor of $a$ and $b$.

We start our treatment of Mordel's work by stating the following:
See proof

This result is already enough for us to make some claims about when $\frac{4}{n}$ has a representation.
For example, suppose $n\equiv -1 \pmod{4}$, i.e., $n$ is of the form $n=4d-1$ for some integer $d$. Then the values $a=2$, $b=1$, and $c=1$ satisfy the conditions of $[\ref{suff1}]$ (verify this for yourself: try plugging in $n=7$). Similarly, when $n\equiv -2 \pmod{4}$ the values $a=b=c=1$ satisfy the conditions of $[\ref{suff2}]$. When $n\equiv 0 \pmod{4}$, we can set $x=y=z=\frac{3n}{4}$.

Here is another way to look at this: in equation ($\ref{suff1}$), If we let $a=2$, $b=1$, and $c=1$, we get the equation \[4(2)(1)(1)d = 2 + n1 + n1,\] or \[8d=2+2n \iff 4d=1+n\] This equation only holds when $n\equiv -1 \pmod {4}$, just as we had suspected! If we continue assigning values to $a$, $b$, and $c$, we can get more congruence relations like this. Here are a couple of examples:

Plug and chug method: Equation ($\ref{suff1}$)
$[a,b,c]$	Simplified equation ($\ref{suff1}$)	Congruence relation?
$[1,1,1]$	$1+2n=4d$	Failure!
$[2,1,1]$	$2+2n=8d$	$n\equiv 3 \pmod{4}$
$[1,2,1]$	$1+3n=8d$	$n\equiv 5 \pmod{8}$
$[1,2,2]$	$1+4n=16d$	Failure!
$[2,2,1]$	$2+3n=16d$	$n\equiv 10 \pmod{16}$
$[2,2,2]$	$2+4n=32d$	Failure!
$[7,9,10]$	$7+19n=2520d$	$n\equiv 2387 \pmod{2520}$

Not that this processes doesnt work all the time: for example, in the case $[1,1,1]$, there is no solution for the equation $1+2n=4d$. Still, it seems like we could get a lot of data out of this, right? Well, it would be redundant for me to tell you "there is a solution when $n$ is a multiple of $4$ AND when $n$ is a multiple of $8$," because if $n$ is a multiple of $8$ then it is necessarily a multiple of $4$. If we remove the redundant entries from our table, we get that $n\equiv 3\pmod{4}$ and $n\equiv 5\pmod{8}$.

Let's try using equation ($\ref{suff2}$) instead. We can re-write equation ($\ref{suff2}$) as \[ c(4abd-1)=na+b \]

Because $4abd-1$ is coprime to $a$, given some positive non-zero numbers $a$, $b$, and $d$, there will always be a soultion to equation ($\ref{suff2}$) This makes filling out our table much easier. Let $k$ be a non-zero number. If we suppose that $abd=k$, then we can see our modulus relations give us information about $n$ modulo $4k-1$. From this, we have the following results for all values of $k$:

• If we set $a=b=1$ and $d=k$, then we get $n\equiv -1 \pmod{(4k-1)}$.
• If we set $b=k$ and $a=d=1$, then we get $n\equiv -k \pmod{(4k-1)}$.
• If we set $a=k$ and $b=d=1$, then we get $n\equiv -4 \pmod{(4k-1)}$.
• By the previous bullet, we have that $n\equiv 0 \pmod{(4k-1)}$ is covered as well (multiples of solutions are covered).

More generally, we can write \begin{equation}\label{cong_suff2}n\equiv -4b^2d\pmod{(4k-1)}\end{equation} for $b$ and $d$ such that $bd$ divides $k$. This gives us the following list:

• $n\equiv 0 \pmod{3}$
• $n\equiv 2 \pmod{3}$
• $n\equiv 0 \pmod{7}$
• $n\equiv 3 \pmod{7}$
• $n\equiv 6 \pmod{7}$
• $n\equiv 5 \pmod{7}$
• $n\equiv 0 \pmod{11}$
• $n\equiv 7 \pmod{11}$
• $n\equiv 10 \pmod{11}$
• $n\equiv 8 \pmod{11}$

and so on and so forth.

Infinite sets of congruence relations

The following are examples of sets of congruence relations:

1. $\{0\pmod{2}, 1\pmod{2}\}$
2. $\{0\pmod{2},1\pmod{4}, 3\pmod{4}\}$
3. $\{0\pmod{2},1\pmod{4}, 3\pmod{8},7\pmod{8}\}$
4. $\{0\pmod{2}\}\cup\{2^k+1\pmod{2^{k+1}}\vert 1\leq k\}$
5. $\{11 \pmod{p} \vert p \text{ prime}\}$

Examples 1, 2, and 3 are finite sets, and examples 4 and 5 are infinite.