# 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:
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