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:


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


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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jul 23 '13 at 13:15
  • $\begingroup$ It would be nice to mention what the conjecture is about, or at least give a link to Wikipedia. $\endgroup$ – Martin Sleziak Jul 23 '13 at 13:33
  • $\begingroup$ I'm sorry, I though it was well known. $\endgroup$ – Jori Jul 23 '13 at 13:39
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    $\begingroup$ Mordell's book "Diophantine Equations" has a section on this. $\endgroup$ – Mike Bennett Jul 23 '13 at 20:55
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    $\begingroup$ 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 -> $p = 3 \bmod 4$. $\endgroup$ – 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.


Then the solutions have the form:




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:


The solutions have the form:




  • $\begingroup$ 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 $q$. $\endgroup$ – Kieren MacMillan Sep 16 '14 at 12:38
  • $\begingroup$ 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. $\endgroup$ – James Arathoon Aug 9 '17 at 10:18

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


$t,q$ - integers.

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

The solutions have the form:




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

The solutions have the form:




  • $\begingroup$ 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. $\endgroup$ – 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 $$\frac1{4N} + \frac1{N/3} + \frac1{4N/3} = \frac4N.$$

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

For the equation.


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.




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



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





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