Erdős–Straus conjecture

Importance: Medium ✭✭
Subject: Number Theory
Keywords: Egyptian fraction
Recomm. for undergrads: yes
Posted by: ACW
on: February 29th, 2012

\begin{conjecture} % Enter your conjecture in LaTeX For all $n > 2$, there exist positive integers $x$, $y$, $z$ such that $$1/x + 1/y + 1/z = 4/n$$. % You may change "conjecture" to "question" or "problem" if you prefer. \end{conjecture}

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: See \Def{Erdős–Straus conjecture} for more details. % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{}


% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

* indicates original appearance(s) of problem.

Formula Individa

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


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

Further restriction

I think you need to specify that $x$, $y$ and $z$ be positive for this to be challenging (and open).


Done. Thank you.

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