<s>[INST] high order integer equation 
Find all tuple $(x,y)$ such that $x,y$ are integers and $(x^2-y^2)^2=20y+1$.

First i see that $x^2-y^2$ is odd and from the fact that a difference between square of two odd is multiple of $8$ and thus $y$ is a multiple of $2$.
Moreover, we have $(x^2-y^2+1)(x^2-y^2-1)=20y$.
Somebody can give some hint! Whether the first information is useful?
 [/INST] Since $x^2-y^2$ is odd then $x\not=y$ and
$$20y+1=(x^2-y^2)^2\geq ((y-1)^2-y^2)^2=(-2y+1)^2=1-4y+4y^2.$$
Hence
$$0\geq 4y^2-24y\Leftrightarrow y(y-6)\leq 0 $$
which implies that $0\leq y\leq 6$.
Moreover $20y+1$ is a perfect square, and therefore $y\in\{0,4,6\}$:

*

*if $y=0$ then  $(x^2-0)^2=1^2$ and $x=\pm 1$;


*if $y=4$ then  $(x^2-16)^2=9^2$ and $x=\pm 5$;


*if $y=6$ then  $(x^2-36)^2=11^2$ and $x=\pm 5$.
Therefore the integer solutions are $(\pm1,0),(\pm5,4),(\pm 5,6).$
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