Erdős–Straus conjecture

Importance: Medium ✭✭

Author(s):	Erdos, Paul
	Straus, Ernst G.

Subject:

Number Theory

Keywords:

Egyptian fraction

Recomm. for undergrads: yes

Posted	by:	ACW
	on:	February 29th, 2012

Conjecture

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.

Bibliography

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Formula Individa

On July 14th, 2014 Anonymous says:

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

Solution

On July 14th, 2014 Anonymous says:

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