# Erdős-Straus conjecture

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.

• I can't exactly point out a summary. But you can check Terence Tao's link which does have some new resources. Do post this on MO. – Torsten Hĕrculĕ Cärlemän Jul 23 '13 at 13:15
• It would be nice to mention what the conjecture is about, or at least give a link to Wikipedia. – Martin Sleziak Jul 23 '13 at 13:33
• I'm sorry, I though it was well known. – Jori Jul 23 '13 at 13:39
• Mordell's book "Diophantine Equations" has a section on this. – Mike Bennett Jul 23 '13 at 20:55
• If you don't want to use p \equiv 3 \pmod{4} ($p \equiv 3 \pmod{4}$), which I'd prefer, you can use \bmod, p = 3 \bmod 4 -> $p = 3 \bmod 4$. – Daniel Fischer Nov 14 '14 at 15:54

For the equation: $$\frac{4}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

The solution can be written using the factorization, as follows.

$$p^2-s^2=(p-s)(p+s)=2qL$$

Then the solutions have the form:

$$x=\frac{p(p-s)}{4L-q}$$

$$y=\frac{p(p+s)}{4L-q}$$

$$z=L$$

I usually choose the number $L$ such that the difference: $(4L-q)$ was equal to: $1,2,3,4$ Although your desire you can choose other.

You can write a little differently. If unfold like this:

$$p^2-s^2=(p-s)(p+s)=qL$$

The solutions have the form:

$$x=\frac{2p(p-s)}{4L-q}$$

$$y=\frac{2p(p+s)}{4L-q}$$

$$z=L$$

• Nice formulae! If you can prove that $4L-q$ can always be chosen such that $x$ and $y$ are integral, you should write this up and publish it. A rather amazing extension would be to show how many distinct integer triples $(x,y,L)$ can be obtained for a given $q$ using your formulas, especially if you can also prove that number is maximal for fixed $q$. – Kieren MacMillan Sep 16 '14 at 12:38
• Using this approach can you find the complete subset of positive integers for q, for which this algebraic formula does not work e.g. $q=193$. If you can't then this approach has little value as the starting point for solving this problem. – James Arathoon Aug 9 '17 at 10:18

It was necessary to write the solution in a more General form:

$$\frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$t,q$ - integers.

Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=2qL$

The solutions have the form:

$$x=\frac{p(p-s)}{tL-q}$$

$$y=\frac{p(p+s)}{tL-q}$$

$$z=L$$

Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=qL$

The solutions have the form:

$$x=\frac{2p(p-s)}{tL-q}$$

$$y=\frac{2p(p+s)}{tL-q}$$

$$z=L$$

• This more general form appears only to be relevant to this question when $t=4$. Which immediately takes us back to your previous post. So unless further explanation can be given for the applicability of the generalisation here, I am assuming that comments made in regard to your previous post apply here as well. – James Arathoon Aug 9 '17 at 10:26

When $N$ is pair, then the solution is $$\frac1N + \frac{1}{N/2} + \frac1N = \frac4N.$$

When $N$ is a multiple of $3$, then the solution is $$\frac1{4N} + \frac1{N/3} + \frac1{4N/3} = \frac4N.$$

• – Shaun Feb 16 '15 at 13:47
• @Koussay: The approach used by Mordell (and described above in the newly revised question) includes these two results and much more. – James Arathoon Aug 10 '17 at 12:31

For the equation.

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{t}{q}$$

All variations of the same formula. As the number of solutions of course need to consider all possible factorization. Too much ends quickly. The number must be greater than 1.

$$z=L$$

$$x=\frac{qL(p+s)}{s(tL-q)}$$

$$y=\frac{qL(p+s)}{p(tL-q)}$$

Consider this example. $t=4$ ; $q=193$

$4L=193+i$

$i=3;7;11.....$

Let $i=7$ Means $L=50$ ; $p=193$ ; $s=10$

$z=50$

$x=1450=50*29$

$y=5*29*193=27985$