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I'm reading a lot about the Erdös-Straus Conjecture (ESC), a conjecture that states that for every natural number $p \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:

$$\frac{4}{p}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

(see also: http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture)

In my textbook the conjecture is proved for the following cases:

$$p=3\mod4 \\ p=2\mod3$$

But I know ESC has also been proven for: $$p=2\mod5\\ p=3\mod5\\ p=3\mod7\\ p=5\mod7\\ p=6\mod7\\ p=5\mod8$$

I also know these proves are more difficult, but I'm just curious where I can find a summary of them all.

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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
    
Off topic: does anyone know how to remove that large space between the number and $mod$? –  Jori Jul 24 '13 at 20:08

4 Answers 4

For all $n\geq2$, there exists $x,y,z$ positive integers such that $\cfrac{4}{n}=\cfrac{1}{x}+\cfrac{1}{y}+\cfrac{1}{z}$ if and only if $n$ has no common factor with $\varphi$ and

$$\cfrac{4}{n}=\cfrac{\log(\pi^{\varphi})}{\log(\sqrt{\pi^{\frac{n}{2}\varphi}})},$$

where $\pi$ is any constant$>1$ and $\varphi$ the sum of $x,y,z.$

The $K$ case is easy to prove since one of the integers $x,y,z=1=\cfrac{1}{2}+\cfrac{1}{2}$ such that $K=2.$

See more the article here

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Is there a particular reason you chose to use $\pi$ in your equation, as opposed to anything else? –  Foo Barrigno Mar 6 at 19:28
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@FooBarrigno $\pi$ can be used to represent the means of the complexes numbers $(s)$ as stated by Riemann. It's also the tools which will be used to plot the sequence but there is no other reason in particular. –  user133569 Mar 7 at 15:43
    
rewriting your equation you have $\frac{4}{n}=\frac{\phi*log(\pi)}{n/4*\phi*log(\pi)}=\frac{1}{n/4}=\frac{4}{n}$, so your choice of values of both $\pi$ and $\phi$ seem a little arbitrary. The equation would still be vacuously true if you replaced them with any positive integer. In particular, your equation does not depend on the value of $\phi$ –  Foo Barrigno Mar 7 at 16:28
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@FooBarrigno That is to say since $\pi$ is any constant greater than $1$. It follows that you can replace it by anything else. In this article you should keep in mind the sens that everything comes up from the Riemann theorem. So trying to keep a same line argument makes sense in this context right. Move forward and don't forget the Gauss quotation at the end: It's about harmony, beauty and poetry... –  user133569 Mar 7 at 16:44

For all $n>1$ there exists $x,y,z$ positive integers such that $$\frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ if and only if $$4=\frac{(xyz)^2}{(n?)^2},$$ and $n$ cannot divide the sum of $x,y,z.$

For the Erdos-Straus Triples family we have both major set on the form $$2x=z?$$ and $$x=z?$$

More see here

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this is an update of the comments from user133569 –  Thierno M. SOW Aug 1 at 18:49
    
I don't understand. Why all repeat 100 times ? There is a formula. Substitute it in Your expression. All right reduced and get the result. –  individ Aug 2 at 8:31

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:

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

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

$$z=L$$

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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 at 12:38
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@KierenMacMillan You do these formulas publish. There is enough to consider the division by 3. You can always choose such numbers in order. $L=aL$ ; $q=aq$ ; $a(4L-q)=3a$ $4L-q=a(4L-q)$ ; $(p-s)(p+s)=aL*2aq=aL*2a(4L-3)$ Get: $p=\frac{3a(3L-2)}{2}$ You can always get this number is divisible by 3. –  individ Sep 17 at 6:03

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:

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

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

$$z=L$$

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