Erdős–Straus conjecture

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

For all $ n > 2 $, there exist positive integers $ x $, $ y $, $ z $ such that $$1/x + 1/y + 1/z = 4/n$$.

See Erdős–Straus conjecture for more details.


* 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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