Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.

Yet an other answer following the same natural lines.
The cases when $x=0$ and/or $y=0$ are clear from the OP. We search for other, "new" solutions.
Let $a=(x,y)$ be the lcm of $x,y$, taken now to be $\ge 1$,
so we can and do write $x=aX$, $y=aY$ with relatively prime integers $X,Y$.
Then the given equation is equivalent to:
$$
a(X+Y)(X^2-XY+Y^2) = X^2+42XY+Y^2\ .
$$
This implies:

*

*$(X+Y)$ divides $40XY=(X^2+42XY+Y^2)-(X+Y)^2$.
So $\color{brown}{(X+Y)}$ divides $\color{brown}{40}$.


*$(X^2-XY+Y^2)$ divides $43XY=(X^2+42XY+Y^2)-(X^2-XY+Y^2)$.
So $\color{brown}{(X^2-XY+Y^2)}$ divides $\color{brown}{43}$.
(If a prime $p$ divides $(X+Y)$ and $XY$, then it divides either $X$ or $Y$, possibly exchanging notations let $p$ divide $X$, the $p$ divides also $(X+Y)-X=Y$, so $p$ divides the lcm $(X,Y)=1$, contradiction. The other case is similar.)
From here, some few cases have to be checked:

*

*$X+Y=d\in\{\pm 1,\pm 2,\pm 4,\pm 5,\pm 8,\pm 10,\pm 20,\pm 40\}$, and

*$X^2-XY+Y^2\in\{1,43\}$.

The first restriction lets us linearly substitute $Y$ in terms of $X$. The substitution
$Y=d-X$, inserted in the second restriction gives $3X^2-3dX+d^2\in\{1,43\}$.
thus getting some few equations of second degree in $X$ of the shape:
$$
\begin{aligned}
3X^2 - 3dX + (d^2-1) &=0\ ,\\
3X^2 - 3dX + (d^2-43) &=0\ .
\end{aligned}
$$
We obtain $(X,Y)$-solutions iff the discriminant is a square. So either
$(4-d^2)$, or
$(172-d^2)$ is a square.
Only the values $d=\pm 1$ and $d=\pm 2$ make the first expression ($1^2$ and $0^2$) a square.
Only the values $d=\pm 5$ and $d=\pm 8$ make the second expression ($7^2$ and $6^2$) a square.
We may and do assume $d>0$ (possibly changing now $a$ into $-a$), since passing from $d$ to $-d$ means replacing $(X,Y)$ by $(-X,-Y)$ (and $a$ by $-a$), which leads finally to the same final solution.
Here are these four cases explicitly displayed, and the four final solutions.

*

*$d=1$ :: The corresponding case is $Y=1-X$ with $3X^2-3X+1=1$. So $X\in\{0,1\}$. No "new" solutions.

*$d=2$ :: The corresponding case is $Y=2-X$ with $3X^2-6X+4=1$. So $X=1=Y$. The new solution is $(X,Y)=(1,1)$. It leads to $a=(1+42+1)/(1+1)=22$, and to the corresponding final solution $\color{blue}{(x,y)=(22,22)}$.

*$d=5$ :: The corresponding case is $Y=5-X$ with $3X^2-15X+25=43$.
The new solutions are $(X,Y)=(6,-1)$, and $(X,Y)=(-1,6)$.
The corresponding value of $a$ is $(36-6\cdot 42+1)/(216-1)=-1$.
So the final solutions are $\color{blue}{(x,y)=(-6,1)}$, and $\color{blue}{(x,y)=(1,-6)}$.

*$d=8$ :: The corresponding case is $Y=8-X$ with $3X^2-24X+64=43$.
The new solutions are $(X,Y)=(7,1)$, and $(X,Y)=(1,7)$.
The corresponding value of $a$ is $(49+7\cdot 42+1)/(343+1)=1$, leading to the copies $\color{blue}{(x,y)=(7,1)}$ and $\color{blue}{(x,y)=(1,7)}$.

