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

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