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 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:

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:

Hence the solution: