# Math Help - Difficult square number problem

1. ## Difficult square number problem

A pair of positive integers $(x, y)$ is called 'square' if both $x + y$and $xy$ are perfect square numbers. E.g. $(5, 20)$ is 'square' since $5 + 20 = 25$ and $5 x 20 = 100$, both which are perfect squares.

Prove no square pair exists in which one of its numbers is 3.

I'm not sure how to do this, but I think it involves mods in the solution.

2. Hello BG5965
Originally Posted by BG5965
A pair of positive integers $(x, y)$ is called 'square' if both $x + y$and $xy$ are perfect square numbers. E.g. $(5, 20)$ is 'square' since $5 + 20 = 25$ and $5 x 20 = 100$, both which are perfect squares.

Prove no square pair exists in which one of its numbers is 3.

I'm not sure how to do this, but I think it involves mods in the solution.

There may be a shorter route to the contradiction, but the following works OK.

Suppose that $(3,x)$ is a square pair. Then, for some $p, q$:

$3+x = p^2$ and $3x=q^2$

$\Rightarrow q$ is a multiple of $3$; say $q = 3r$

Then $3x = q^2 = 9r^2$

$\Rightarrow x = 3r^2$

$\Rightarrow 3 + x = 3+3r^2 = p^2$

$\Rightarrow 3(1+r^2) = p^2$

$\Rightarrow p$ is a multiple of $3$; say $p = 3s$

$\Rightarrow p^2 = 9s^2 = 3(1+r^2)$

$\Rightarrow 1+r^2 = 3s^2$

$\Rightarrow r^2 \equiv 2 \mod 3$

But this is impossible, since $0^2\equiv 0$ and $1^2 \equiv 2^2 \equiv 1 \mod 3$.

Hence the square pair $(3, x)$ does not exist.