The Erdős-Straus Conjecture (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:
$$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$
The basic approach to solving this problem outlined by Mordell [Ref1] is described below
By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that
a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$
In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$
b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus
$$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$
will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.
Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$
The argument continues...
Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5} \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$
Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$
Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n<1009$.
This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.
Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.
The question is:
Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?
[Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290.
p \equiv 3 \pmod{4}
($p \equiv 3 \pmod{4}$), which I'd prefer, you can use\bmod
,p = 3 \bmod 4
-> $p = 3 \bmod 4$. $\endgroup$