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:

\(\displaystyle 41 + 80 + 320\) \(=\) \(\displaystyle 441\)
\(\displaystyle \) \(=\) \(\displaystyle 21^2\)
\(\displaystyle 41 + 80\) \(=\) \(\displaystyle 121\)
\(\displaystyle \) \(=\) \(\displaystyle 11^2\)
\(\displaystyle 41 + 320\) \(=\) \(\displaystyle 361\)
\(\displaystyle \) \(=\) \(\displaystyle 19^2\)
\(\displaystyle 80 + 320\) \(=\) \(\displaystyle 400\)
\(\displaystyle \) \(=\) \(\displaystyle 20^2\)


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:

\(\displaystyle p\) \(=\) \(\displaystyle \paren {p + q + r} - \paren {q + r}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {x^2 + 2 x + 1} - \paren {x^2 - 2 x + 1}\) multiplying out the expression for $q + r$
\(\displaystyle \) \(=\) \(\displaystyle 4 x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle q\) \(=\) \(\displaystyle x^2 - 4 x\)

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:

\(\displaystyle p\) \(=\) \(\displaystyle 80\)
\(\displaystyle q\) \(=\) \(\displaystyle 320\)
\(\displaystyle r\) \(=\) \(\displaystyle 41\)

$\blacksquare$


Sources