Find all integral solutions to $xy^2 + (x^2 + 1)y + x^4 + 1 = 0$. I have an equation $$xy^2 + (x^2 + 1)y + x^4 + 1 = 0$$ and I need to find a set of solutions for it in integers. I tried to solve it using Maple with the isolve () command, but it doesn’t give anything. This suggests that either the Maple algorithms cannot solve it or that there are no solutions. However, I know that he is resolved. Maybe you know special algorithms for solving this equation or other methods?
 A: This answer makes use of the rational root theorem a lot. Let $x$ and $y$ be integers such that
$$xy^2 + (x^2 + 1)y + x^4 + 1 = 0.\tag{1}$$
We first make a few simple observations.


*

*If $x=0$ then $y=-1$, and if $y=-1$ then $x=0$.

*There are no other integral solutions with $|x|\leq2$.

*Equation $(1)$ shows that $y$ divides $x^4+1$. 


Lemma 1: If $(x,y)$ is an integral solution to $(1)$, then there exists $u\in\Bbb{Z}$ such that
$$y=ux^4-2x^3-x^2-x-1.$$
Proof. If $x=0$ then this follows from the first observation above. So suppose $x\neq0$. Viewing $(1)$ as a polynomial in $x$ we see that
$$x^4+yx^2+y^2x+y+1=0,$$
and hence that $x$ divides $y+1$. Let $z=\tfrac{y+1}{x}\in\Bbb{Z}$ so that plugging in $y=zx-1$ into $(1)$ yields
$$x^3+x^2z^2+x^2z-2xz-x+z+1=0.$$
this in turn shows that $x$ divides $z+1$. Let $w=\tfrac{z+1}{x}\in\Bbb{Z}$ so that plugging in $z=wx-1$ yields
$$w^2x^3-wx^2-2wx+w+x^2+1=0,$$
which in turn shows that $x$ divides $w+1$. Let $v=\tfrac{w+1}{x}\in\Bbb{Z}$ so that plugging in $w=vx-1$ yields
$$v^2x^4-2vx^3-vx^2-2vx+v+x^2+2x+2=0.$$
which in turn shows that $x$ divides $v+2$. Let $u=\tfrac{v+2}{x}\in\Bbb{Z}$ so that
\begin{eqnarray*}
y&=&zx-1=(wx-1)x-1=((vx-1)x-1)x-1\\
&=&(((ux-2)x-1)x-1)x-1\\
&=&ux^4-2x^3-x^2-x-1.\hspace{100pt}\square
\end{eqnarray*}
Lemma 2: If $x\neq0$ then $|u|\leq1$.
Proof. If $|u|>1$ then for all $x\notin\{-1,0,1,2\}$ we have
$$|y|=|ux^4-2x^3-x^2-x-1|>x^4+1,$$
contradicting the fact that $y$ divides $x^4+1$. As there are no integral solutions with $x\in\{-1,1,2\}$ it follows that $|u|\leq1$ if $x\neq0$.$\hspace{20pt}\square$
At this point we can simply try the three possible values of $u$ to get three polynomials in $x$, and see whether they have integral roots:
If $u=0$ then $y=-(2x^3+x^2+x+1)$. Plugging this back into $(1)$ yields
$$x^4(4x^3+4x^2+3x+6)=0,$$
and the rational root theorem shows that $x=0$ is the unique solution.
If $u=1$ then $y=x^4-2x^3-x^2-x-1$. Plugging this back into $(1)$ yields
$$x^4(x^2+1)(x^3-4x^2+x+7)=0,$$
which again has $x=0$ as its unique integral solution by the rational root theorem.
If $u=-1$ then $y=-x^4-2x^3-x^2-x-1$. Plugging this back into $(1)$ yields
$$x^4(x^5+4x^4+6x^3+5x^2+5x+5)=0,$$
which again has $x=0$ as its unique integral solution by the rational root theorem.
This shows that $(x,y)=(0,-1)$ is the unique integral solution to $(1)$.
