# 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

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

where $u$ and $v$ are integers such that $uv$ 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:

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

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

 $\displaystyle x$ $\displaystyle=$ $\displaystyle mv(u+v)$ $\displaystyle y$ $\displaystyle=$ $\displaystyle mu(u+v)$

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

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

Title solution of $1/x+1/y=1/n$ SolutionOf1x1y1n 2013-03-22 16:30:44 2013-03-22 16:30:44 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Theorem msc 11D99