Diophantus of Alexandria/Arithmetica/Book 5/Problem 30

From ProofWiki
Jump to navigation Jump to search

Example of Diophantine Problem

A man buys a certain number of measures of wine, some at $8$ drachmas, some at $5$ drachmas each.
He pays for them a square number of drachmas;
and if we add $60$ to this number, the result is a square, the side of which is equal to the whole number of measures.
Find how many he bought at each price.


Solution

The solution given by Diophantus of Alexandria is:

The number of $5$-drachma measures is $\dfrac {79} {12}$.
The number of $8$-drachma measures is $\dfrac {59} {12}$.


Proof

Let $x$ be the total number of measures bought.

Then $x^2 - 60$ is the price paid, which is itself a square number: $\paren {x - m}^2$, say.

So, $\dfrac 1 5$ of the price of the $5$-drachma measures plus $\dfrac 1 8$ of the price of the $8$-drachma measures equals $x$.

Hence the total price $x^2 - 60$ has to be divided into $2$ parts such that $\dfrac 1 5$ of one plus $\dfrac 1 8$ of the other equals $x$.

This has no real solution unless:

\(\ds x\) \(>\) \(\ds \dfrac 1 5 \paren {x^2 - 60}\)
\(\ds x\) \(<\) \(\ds \dfrac 1 8 \paren {x^2 - 60}\)
\(\ds \leadsto \ \ \) \(\ds 5 x\) \(<\) \(\ds x^2 - 60 < 8 x\)

Thus we have:

$x^2 > 5 x + 60$

and so $x^2$ equals $5 x$ plus a number greater than $60$.

Hence $x$ is not less than $11$.

Similarly, we have:

$x^2 < 8 x + 60$

and so $x^2$ equals $8 x$ minus a number less than 60.

Hence $x$ is not greater than $12$.

That is:

$11 < x < 12$


From above, we deduce:

$x = \dfrac {m^2 + 60} {2 m}$

and so:

$22 m < m^2 + 60 < 24 m$


So:

$22 m$ equals $m^2$ plus some number less than $60$

and so:

$m^2$ is not less than $19$.

Also:

$24 m$ equals $m^2$ minus some number greater than $60$

and so:

$m^2$ is less than $21$.

So we can put $m = 20$, and so:

$x^2 - 60 = \paren {x - 20}^2$

which leads to:

\(\ds x\) \(=\) \(\ds 11 \tfrac 1 2\)
\(\ds x^2\) \(=\) \(\ds 132 \tfrac 1 4\)
\(\ds x^2 - 60\) \(=\) \(\ds 72 \tfrac 1 4\)


It remains to divide $72 \frac 1 4$ into $2$ parts such that $\dfrac 1 5$ of one part plus $\dfrac 1 8$ of the other part equals $11 \frac 1 2$.

Let the first part be $5 z$.

Therefore $\dfrac 1 8$ times the second half equals $11 \frac 1 2 + z$

That is, the second half equals $92 - 8 z$m which leads to:

$5 z + 92 - 8 z = 72 \frac 1 4$

and so:

$z = \dfrac {79} {12}$

Hence the result.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Diophantus_of_Alexandria/Arithmetica/Book_5/Problem_30&oldid=546668"