# 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:

$p q + r$ is square
$q r + p$ is square
$r p + q$ is square

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 $q = 7$.

$\blacksquare$