$\frac{2019(x^{4}+2)}{x^{2}y+1} \in \mathbb{N^*}$ Find all pair $(x;y)$ of positive integers such that $k=\frac{2019(x^{4}+2)}{x^{2}y+1} \in \mathbb{N^*}$
$\Leftrightarrow 2019x^4-kyx^2+4038-k=0$.Hence $\Delta\geqslant 0$
$\Leftrightarrow k^2y^2+4 \cdot 2019(k-4038)\geqslant 0\Leftrightarrow y^2\geqslant \frac{4 \cdot 2019(4038-k)}{k^2}$
$$x^2y+1 \mid 2019(x^2-2y)$$
I don't know what I should do next. Help me, please?
$2019=3\times 673$
So k must be a multiple of $3$ or $673$
Let $y=tx$ we have:
$2019x^4+4038-ktx^3-k=0$
$x^3(2019 x- kt)=k-4038$
Suppose $k=4038$ we have:
$2019 x= 4038 t$
$(t, x, y)=(1, 2, 2), (2, 4, 8), (3, 6, 18), . . .$
Let $k=673$ we get:
$x^3(2019 x -kt)=-3365$
For $x=1$ we get:
$2019-kt=-3365$
Which gives $t=8$ and $y=8$ so the couple is $(x, y)=(1, 8)$
You may get some more family of solutions for other multiples of $673$ for k.
$2019=3\times 673$
So k must be a multiple of $3$ or $673$
Let $y=tx$ we have:
$2019x^4+4038-ktx^3-k=0$
$x^3(2019 x- kt)=k-4038$
Suppose $k=4038$ we have:
$2019 x= 4038 t$
$(t, x, y)=(1, 2, 2), (2, 4, 8), (3, 6, 18), . . .$
Let $k=673$ we get:
$x^3(2019 x -kt)=-3365$
For $x=1$ we get:
$2019-kt=-3365$
Which gives $t=8$ and $y=8$ so the couple is $(x, y)=(1, 8)$
You may get some more family of solutions for other multiples of $673$ for k.