# Erdös-Straus conjecture

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:

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

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

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 '14 at 12:38
@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 '14 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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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 $$\frac1{4N} + \frac1{N/3} + \frac1{4N/3} = \frac4N.$$

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–  Shauna Feb 16 at 13:47