<s>[INST] The equation with integer solution: $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$ I am trying to solve this equation:
Given $x,y\in \text {Z}$: Solve : $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$
Obviously, when $x=0$, it is easy to find that $y=0,y=1$ and $y=0$ then $x=0, x=-1$.
My question is how to find another solutions.
My try is factorization, but the power of $x,y$ is too big, i guess there is relation between $x,y$ so i tried to divide polynomial $x^4+x-y^4+y-2x^2y^2(x^4-y^4+1)^3$ by $xy$ or $xy-1$ (base on CAS gives solutions $x,y$ and $xy=0,xy=1$) but i failed.
I am bad at inequalities so i can't find another approach.
Every help is precious to me,thank you for reading.
 [/INST] Write $a=x+y$, $b=x-y$, $c=x^2+y^2$ then the equation becomes
$$2a(bc+1)=(c-ab)(c+ab)(abc+1)^3.$$
Now if $|a|>1$ and $|bc|>1$ then by the fundamental theorem of arithmetic either $a$ or $bc+1$ must contain three copies of all the prime factors of $abc+1$. However $a$ is coprime with $abc+1$, so  $bc+1$ must contain three copies of all the prime factors of $abc+1$, an impossibility. Hence there are no integer solutions other than the trivial cases.
★ Note that
$$
\begin{align}
c-ab&=x^2+y^2-(x^2-y^2)=2y^2\\
c+ab&=x^2+y^2+(x^2-y^2)=2x^2\\
abc&=x^4-y^4\\
\end{align}
$$
</s>