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:

$\ds p$

$=$

$\ds 80$

$\ds q$

$=$

$\ds 320$

$\ds r$

$=$

$\ds 41$

$\blacksquare$

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