Erdős-Straus conjecture

Question

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.

Answer 1

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:

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

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

$$z=L$$

Answer 2

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

Answer 3

"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).

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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.

Answer 4

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).

Answer 5

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 $-a<k$, then the statement holds.