# 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

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

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

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

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

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

∎

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 |