Diophantus of Alexandria/Arithmetica/Book 3/Problem 6
Jump to navigation
Jump to search
Example of Diophantine Problem
To find $3$ numbers such that their sum is a square and the sum of any pair of them is a square.
That is, let $\set {p, q, r}$ be a set of $3$ natural numbers such that:
What are those $3$ numbers?
Solution
The solution given by Diophantus of Alexandria is:
- $\set {41, 80, 320}$
As can be seem:
\(\ds 41 + 80 + 320\) | \(=\) | \(\ds 441\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 21^2\) | ||||||||||||
\(\ds 41 + 80\) | \(=\) | \(\ds 121\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 11^2\) | ||||||||||||
\(\ds 41 + 320\) | \(=\) | \(\ds 361\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 19^2\) | ||||||||||||
\(\ds 80 + 320\) | \(=\) | \(\ds 400\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 20^2\) |
Proof
Let $p + q + r$ be $x^2 + 2 x + 1$.
Let:
- $p + q = x^2$
and so:
- $r = 2 x + 1$
Let $q + r = \paren {x - 1}^2$.
Therefore:
\(\ds p\) | \(=\) | \(\ds \paren {p + q + r} - \paren {q + r}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x^2 + 2 x + 1} - \paren {x^2 - 2 x + 1}\) | multiplying out the expression for $q + r$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds x^2 - 4 x\) |
But we have that $p + r$ is a square.
That is:
- $6 x + 1 = m^2$
which is satisfied by $m^2 = 121$.
This gives us:
- $x = 20$
and so the numbers are:
\(\ds p\) | \(=\) | \(\ds 80\) | ||||||||||||
\(\ds q\) | \(=\) | \(\ds 320\) | ||||||||||||
\(\ds r\) | \(=\) | \(\ds 41\) |
$\blacksquare$
Sources
- c. 250: Diophantus of Alexandria: Arithmetica: Book $\text {III}$: Problem $6$
- 1910: Sir Thomas L. Heath: Diophantus of Alexandria (2nd ed.): The Arithmetica: Book $\text {III}$: $6$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Square Problems: $28$
- 2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $7$: Patterns in Numbers: Diophantus