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.

COMMENT.-For $x=y$ one has $2x^3=44x^2$ so $x=0$ and $x=22$. Besides we have
$(x+y)^3-3xy(x+y)=(x+y)^2+40xy\iff S^3-3SP=S^2+40P$ where $S$ is sum and $P$ is product. Note that if $(x,y)$ is solution so is $(y,x)$.
We have $S|40P$, and $P=\dfrac{S^3-S^2}{3S+40}$. We consider first $S|40$ so $S=1,2,4,5,8,10,20,40$.
$$S=1\Rightarrow 43P=0\Rightarrow (x,y)=(1,0)\\S=2\Rightarrow4=46P\\S=4\Rightarrow P=\frac{48}{52}\\S=5\Rightarrow P=\frac{100}{55}\\S=8\Rightarrow P=\frac{448}{64}=7\Rightarrow X^2-8X+7=0\Rightarrow (x,y)=(7,1)$$
$S=10,20,40$ gives $P=\dfrac{900}{70},76,390$ respectively and the two last integers give $x=10\pm\sqrt{24}$ and $x=20\pm\sqrt{10}$.It remains to look at other possibilities
Thus we have found the solutions $(0,0),(22,22),(1,0),(7,1)$
given above by other users. The other solution is given taking $S=-5$ which gives $P=\dfrac{-150}{25}=-6$ from which the equation $X^2+5X-6=0$ so $(x,y)=(-6,1)$.
Here we want just to give another way of calculation.
