# 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. Jul 23, 2013 at 13:15
• It would be nice to mention what the conjecture is about, or at least give a link to Wikipedia. Jul 23, 2013 at 13:33
• I'm sorry, I though it was well known.
– Jori
Jul 23, 2013 at 13:39
• Mordell's book "Diophantine Equations" has a section on this. Jul 23, 2013 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 -> $p = 3 \bmod 4$. Nov 14, 2014 at 15:54

## 5 Answers

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 $q$. Sep 16, 2014 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. Aug 9, 2017 at 10:18
• The problem with this is that it is not easily verifyable. You did not rule out non-prime numbers. If only primes are investigated, then it is easy to see that $q = 2$ is a special case, and ruling it out you cannot have $(4L - q) = 2$. You can also not say something like "You can write a little differently": at the first factorisation you have $2qL$, which means if q is an odd prime, then depending on the parity of L you may have one more factor of 2. As the parity of $(p - s)$ and $(p + s)$ is the same, the LHS has zero or $2^{\alpha}$ factors of 2. Nov 19, 2020 at 17:25
• Therefore if L is odd, then you must use $qL$ on the RHS to have zero factor of 2, otherwise you have only one but that is invalid. So you must use it, it is not a matter of desire. The same for the value of $(4L - q)$, how could one use any number of desire that is not 1 and still be able to divide both numerators? Nov 19, 2020 at 17:25
• There are only 6 possible triplets in the case of $4/193$ according to maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/… The solution you gave for $4/193$ was using a significantly different formula (without factorisation of a difference of two squares) in another answer. Nov 24, 2020 at 17:05

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 $$\frac1{4N} + \frac1{N/3} + \frac1{4N/3} = \frac4N.$$

• Feb 16, 2015 at 13:47
• @Koussay: The approach used by Mordell (and described above in the newly revised question) includes these two results and much more. Aug 10, 2017 at 12:31

"Are there any other elementary approaches to solving this problem..." - this question is not asked properly, as this problem is not solved, only partially, as to my best knowledge (I mean I see this question is old, but it seems to also be true now). So if the question is about how to find residue classes other than Mordell's, but still not covering all primes thus not proving the conjecture, then I can show that the following $$n$$'s are all possible to be expanded into 3 terms $$\forall p, x \in \mathbb{Z}^+$$:
$$n = (4p - 1)x - 1$$
$$n = (4p - 1)x - p$$
The smallest prime number that is not covered by Mordell's residue classes is $$4201$$ that can be produced with the former, and is $$1009$$ with the latter.
You can see the identity here, in Part IV. - II. (little after the middle).

Update:
I attributed the above identities to myself, but Mordell would possibly kick me in the butt: he got all the residue classes from the condition $$na + b + c = 4abcd$$.
If I let $$a = b = 1$$, then $$n = (4d - 1)c - 1$$, and
if I let $$a = d = 1$$, then $$n = (4c - 1)b - c$$, so the above can be produced from his condition, they are just not part of the usual residue class analyses, I guess because they are hard to handle.

This is a description of an approach to the problem I have not been able to get to work, but nevertheless may be of interest to others.

If one wants to study the Erdős–Straus conjecture there there are many examples of formulas giving Egyptian Fraction Triple Solutions given a fraction $$\frac{4}{n}$$ in the mathematical literature and on this site for example.

These formulas come with various numbers of free parameters and often have divisibility conditions/constraints associated with them that does not always guarantee correct solutions; especially in the case of the residue classes targeted by Mordell and of most interest in regard to the Erdős–Straus conjecture, that I have listed in the revised question above, for example $$n \equiv 1 \pmod{4}$$.

However I devised a formula (in 2016/17 when I became interested in Egyptian fractions), with 3 integer parameters, with no nasty divisibility conditions that lead to false solutions (non-triples), and that works with a selected residue class $$n \equiv 1 \pmod{8}$$ or $$n \equiv 1 \pmod{24}$$ for example.

$$\frac{4}{8 t+1}=\frac{1}{2 t+r}+\frac{1}{m (8 t+1) (2 r+a)}+\frac{1}{m (8 t+1) (2 r-a-1)} \tag{1}$$

Positive numbers, $$8t+1$$, in the desired residue class can be generated using a linked formula such as

$$8t+1=16mr^2 -4r(2m+1) -4m(a^2+a) +1 \tag{2}$$

(combinations of the 3 positive integer parameter values are not allowed when they give a negative number result).

As pointed out by Dávid Laczkó, $$r$$ is always even so if $$r=2s$$ (2) becomes

$$8t+1=64ms^2 -8s(2m+1) -4m(a^2+a) +1 \tag{3}$$

An allowed combination of the three parameters will always result in a number in the desired residue class, as well as an associated Egyptian Fraction Triple Solution.

If it could be proved that formula (2) has full coverage of the applicable residue class(es) to the problem, then this might provide an alternative way of proving the Erdős–Straus conjecture, given the direct one-to-one mapping between formulas (1) and (2) (given identical input parameter values).

Since the numbers $$n$$ are not neatly ordered along a 1-D number line, but ordered in a much more difficult fashion across a portion of a 3-D number space, it is difficult (for me) to find a mathematical proof that, at the very least, all the prime numbers in the selected residue class are there in the set of numbers generated by (2).

(Sorry if I'm not using the correct mathematical terminology to describe this, in the last paragraph above).

• Is $(1)$ an identity or a heuristic formula that happens to work? I could not do the transformations to get an identity, neither some online algebraic expression simplifiers. What I could verify is that for parity reasons, $r$ must always be even. Nov 25, 2020 at 13:48
• Its a data driven heuristic formula that seems to work. I studied the form of triples for 4/p primes with small numbers of solutions and looked for a form of solution that occurred regularly and that I could make sense of. Yes your right $r$ is always even. (2) is a quadratic in $r$, so $r$ can be calculated in terms of $m$, $a$ and $t$. Nov 25, 2020 at 14:15

As a note, when $$p$$ is a prime of the form $$p=4a^2+2a-1$$, the conjecture holds. More generally, if $$p$$ is of the form $$p=4a^2+4ak-2a-k$$ where $$a$$ is a positive integer $$k$$ is another integer satisfying $$-a, then the statement holds.

• Given 11 users' answers, aim your talent to newer, or unanswered questions. (a little overlap of hours, days, or weeks is fine). The question has already been answered, eight years ago. Not a bad idea for practice, but given your handle on formatting, and your argument and fine answer, I encourage you to answer more recent questions. Good luck! May 10, 2022 at 18:26