solution of $1/x+1/y=1/n$

Theorem 1.

Given an integer $n$, if there exist integers $x$ and $y$ such that

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{n},$$

then one has

$x$ $=$ ${\displaystyle \frac{n(u+v)}{u}}$

$y$ $=$ ${\displaystyle \frac{n(u+v)}{v}}$

where $u$ and $v$ are integers such that $u\mathit{}v$ divides $n$.

Proof.

To begin, cross multiply to obtain

$$xy=n(x+y).$$

Since this involves setting a product equal to another product, we can think in terms of factorization. To clarify things, let us pull out a common factor and write $x=kv$ and $y=ku$, where $k$ is the greatest common factor and $u$ is relatively prime to $v$. Then, cancelling a common factor of $k$, our equation becomes the following:

$$kuv=n(u+v)$$

This is equivalent to

$$uv\mid n(u+v)$$

Since $u$ and $v$ are relatively prime, it follows that $u$ is relatively prime to $u+v$ and that $v$ is relatively prime to $u+v$ as well. Hence, we must have that $uv$ divides $n$,

Now we can obtain the general solution to the equation. Write $n=muv$ with $u$ and $v$ relatively prime. Then, substituting into our equation and cancelling a $u$ and a $v$, we obtain

$$k=m(u+v),$$

so the solution to the original equation is

$x$ $=$ $mv(u+v)$

$y$ $=$ $mu(u+v)$

Using the definition of $m$, this can be rewritten as

$x$ $=$ ${\displaystyle \frac{n(u+v)}{u}}$

$y$ $=$ ${\displaystyle \frac{n(u+v)}{v}}.$

∎

Title	solution of $1/x+1/y=1/n$
Canonical name	SolutionOf1x1y1n
Date of creation	2013-03-22 16:30:44
Last modified on	2013-03-22 16:30:44
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	13
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 11D99