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: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:
By dividing both sides of the first equation by $abcd$, we get $4=\frac{a+nb+nc}{abcd}$. We can then split the fraction on the right into three parts, getting $4= \frac{a}{abcd}+\frac{nb}{abcd}+\frac{nc}{abcd}$. Now we divide both sides by $n$, resulting in $\frac{4}{n}=\frac{a}{nabcd}+\frac{nb}{nabcd}+\frac{nc}{nabcd}$. After simplifying each fraction we get $\frac{4}{n}=\frac{1}{nbcd} + \frac{1}{acd} + \frac{1}{abd}$, as desired.▨
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 remember, we are trying to find out when is there a solution. 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 $4$. (Blah blah blah, $n\equiv 2 \pmod{8}$ implies $n\equiv 2 \pmod{4}$).
Let's
