Diophantus of Alexandria/Arithmetica/Book 3/Problem 6

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:

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

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:

 $\ds p$ $=$ $\ds 80$ $\ds q$ $=$ $\ds 320$ $\ds r$ $=$ $\ds 41$

$\blacksquare$