Task 4/84 All rational solutions $(x ; y)$ of the equation

$$
4 x^{2} y^{2}-4 x^{2} y-x^{2}+4 x=2
$$
Since $x \neq 0$ is obvious, division by $x^{2}$ is possible. Equivalent transformations then yield

$$
\begin{gathered}
4 y^{2}-4 y=1-4 x^{-1}+2 x^{-2} \rightarrow 4 y^{2}-4 y+1=2\left(1-2 x^{-1}+x^{-2}\right) \\
(2 y-1)^{2}=2\left(1-x^{-1}\right)^{2} \rightarrow|2 y-1|=\sqrt{2}\left|1-x^{-1}\right|
\end{gathered}
$$

There are now two possibilities:

1. It is $1-x^{-1}=0$, so $x=1$. Then $2 y-1=0$, so $y=0.5$. Thus, a solution $(x=1 ; y=0.5)$ is found.
2. It is $1-x^{-1} \neq 0$. Then division yields

$$
\frac{|2 y-1|}{\left|1-x^{-1}\right|}=\left|\frac{(2 y-1) x}{x-1}\right|=\sqrt{2}
$$

If $x$ and $y$ were both rational, then the term on the left side of the equation and thus $\sqrt{2}$ would also be rational. However, this contradicts the known fact that $\sqrt{2}$ is an irrational number. Therefore, the pair $(1 ; 0.5)$ is the only rational solution.