Diophantus of Alexandria/Arithmetica/Book 3/Problem 6

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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