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

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# 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:

$kuv=n(u+v)$ |

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

$\displaystyle x$ | $\displaystyle=$ | $\displaystyle mv(u+v)$ | ||

$\displaystyle y$ | $\displaystyle=$ | $\displaystyle mu(u+v)$ |

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

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

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

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## Mathematics Subject Classification

11D99*no label found*

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