# Thread: Difficult square number problem

1. ## Difficult square number problem

A pair of positive integers $\displaystyle (x, y)$ is called 'square' if both $\displaystyle x + y$and $\displaystyle xy$ are perfect square numbers. E.g. $\displaystyle (5, 20)$ is 'square' since $\displaystyle 5 + 20 = 25$ and $\displaystyle 5 x 20 = 100$, 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 $\displaystyle (x, y)$ is called 'square' if both $\displaystyle x + y$and $\displaystyle xy$ are perfect square numbers. E.g. $\displaystyle (5, 20)$ is 'square' since $\displaystyle 5 + 20 = 25$ and $\displaystyle 5 x 20 = 100$, 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 $\displaystyle (3,x)$ is a square pair. Then, for some $\displaystyle p, q$:

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

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

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

$\displaystyle \Rightarrow x = 3r^2$

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

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

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

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

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

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

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

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