Diophantus of Alexandria/Arithmetica/Book 3/Problem 12
Example of Diophantine Problem
To find $3$ numbers such that the product of any $2$ of them added to the $3$rd gives 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:
- $\set {1, 7, 9}$
As can be seem:
\(\ds 1 \times 7 + 9\) | \(=\) | \(\ds 16\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^2\) | ||||||||||||
\(\ds 7 \times 9 + 1\) | \(=\) | \(\ds 64\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8^2\) | ||||||||||||
\(\ds 9 \times 1 + 7\) | \(=\) | \(\ds 16\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^2\) |
Proof
Take a square and subtract part of it for $r$.
Let $x^2 + 6 x + 9$ be $p q + r$, and let $r = 9$.
Hence:
- $p q = x^2 + 6 x = x \paren {x + 6}$
Let $p = x$ so that $q = x + 6$.
By the two remaining conditions:
- $10 x + 54$
- $10 x + 6$
are both square.
Thus we need to find $2$ squares whose difference is $48$.
In the words of Diophantus of Alexandria:
- This is easy and can be done in an infinite number of ways.
This of course is not true, but the squares $16$ and $64$ come directly to mind.
Equating these to the given expressions:
\(\ds 10 x + 54\) | \(=\) | \(\ds 64\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 1\) |
and also:
\(\ds 10 x + 6\) | \(=\) | \(\ds 16\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 1\) |
and so $p = 1$.
The final number $q = x + 6$ yields that $q = 7$.
$\blacksquare$
Sources
- c. 250: Diophantus of Alexandria: Arithmetica: Book $\text {III}$: Problem $12$
- 1910: Sir Thomas L. Heath: Diophantus of Alexandria (2nd ed.): The Arithmetica: Book $\text {III}$: $12$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Square Problems: $27$