# Diophantus of Alexandria/Arithmetica/Book 1/Problem 22/6, 4, 5

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## Example of Diophantine Problem

*To find $3$ numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal.*

*Let:*

*the first give $\dfrac 1 3$ of itself to the second**the second give $\dfrac 1 4$ of itself to the third**the third give $\dfrac 1 5$ of itself to the first.*

*What are the numbers?*

## Solution

The solution given by Diophantus of Alexandria is:

- $\set {6, 4, 5}$

but it is only the ratio which is important, so:

- $\set {12, 8, 10}$

is another solution.

## Proof

Let $m$ be the final number.

Let $x = 3 p$.

$y$ is assumed to be divisible by $4$.

Let $y = 4$ be chosen.

Thus after giving and taking to and from $y$, we have:

\(\ds m\) | \(=\) | \(\ds \paren {4 - 1} + p\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds p + 3\) |

Thus $x$ must also become $p + 3$.

This $x$ must have taken $p + 3 - 2 p$, that is, $3 - p$.

Thus $3 - p$ must also be $\dfrac z 5$.

That is:

- $z = 15 - 5 p$

Thus:

\(\ds 15 - 5 p - \paren {3 - p} + 1\) | \(=\) | \(\ds p + 3\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds 13 - 4 p\) | \(=\) | \(\ds p + 3\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds p\) | \(=\) | \(\ds 2\) |

Hence the solution:

\(\ds x\) | \(=\) | \(\ds 6\) | ||||||||||||

\(\ds y\) | \(=\) | \(\ds 4\) | ||||||||||||

\(\ds z\) | \(=\) | \(\ds 5\) |

$\blacksquare$

## Sources

- c. 250: Diophantus of Alexandria:
*Arithmetica*: Book $\text I$: Problem $22$ - 1910: Sir Thomas L. Heath:
*Diophantus of Alexandria*(2nd ed.): The Arithmetica: Book $\text {I}$: $22$ - 1992: David Wells:
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