Show that given any rational number $x$, there exists an integer $y$ such that $x^2y$ is an integer

Question

I have a question:

Show that given any rational number $x$, there exists an integer $y$ such that $x^2y$ is an integer.

I am new to proofs, but I have the following so far:

Definitions:

$x$ is a Rational number

$y$ is an Integer

$x^2y$ is an Integer

Assume that $ x = \frac{m}{n}$ where $m$ is an Integer and $n$ is an integer and $n\neq 0$

$x^2 = \frac{m^2}{n^2}$

$x^2y = \frac{m^2}{n^2}y$

I have no idea how to go beyond this to prove that $x^2y$ is an Integer in proof form. I think it is that $\frac{m^2y}{n^2}$ should not have a remainder, but how do I write it in proof form?

Answer 1

Write $x=m/n$ with integers $m$ and $n,$ and then choose $y=n^2.$

Then we have $$ x^2 y = \frac{m^2}{n^2} \cdot n^2 = m^2, $$ which is an integer.