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

## 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:

 $\ds x$ $=$ $\ds 6$ $\ds y$ $=$ $\ds 4$ $\ds z$ $=$ $\ds 5$

$\blacksquare$