Diophantus of Alexandria/Arithmetica/Book 1/Problem 22

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.

That is:

Let $\dfrac 1 p, \dfrac 1 q, \dfrac 1 r$ be given.

The exercise is to find a set of $3$ natural numbers $\set {x, y, z}$ such that:

\(\ds x - \frac x p + \frac z r\)

\(=\)

\(\ds m\)

\(\ds y - \frac y q + \frac x p\)

\(=\)

\(\ds m\)

\(\ds z - \frac z r + \frac y q\)

\(=\)

\(\ds m\)

where $m$ is a natural number.

Example: $\dfrac 1 3, \dfrac 1 4, \dfrac 1 5$

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?

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: Curious and Interesting Puzzles ... (previous) ... (next): The First Pure Number Puzzles: $26$