<s>[INST] Integer equations I have $2$ following problems. Find integer roots of
$$\begin{align}
&1)~\frac{x+y}{x^2-xy+y^2}=\frac3z \\
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
\end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $\left( -2, 1\right), \left( 0, -1\right), \left( 0, 3\right), \left( 2, 1\right) $. Please help me. Thank you very much.
 [/INST] Question 1:
\begin{align}
& \frac{x+y}{x^2-xy+y^2} = \frac 3z \\
\iff & \frac{3(x^2-xy+y^2)}{x+y} = z\\
\iff & 3(x+y) - \frac{9xy}{x+y} = z \\
\iff & \frac{9xy}{x+y} = 3x+3y-z \\
\implies & \frac{9}{\frac 1x + \frac 1y} = 3x+3y-z
\end{align}
From the LHS we see that $\frac 1x + \frac 1y$ must be equal to $\pm 1, \pm 3, \pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $\frac 1x + \frac 1y \in [-2,2]$ and it follows that $\frac 1x + \frac 1y = \pm 1$.
Moreover, the only way this can happen is if $x = y = \pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $\iff$ to an $\implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t \in \Bbb Z$ with $t \neq 0$.
</s>