Solve $(y^2 + xy)(x^2 - x + 1) = 3x - 1$ over the integers. 
Solve $$(y^2 + xy)(x^2 - x + 1) = 3x - 1$$ over the integers.

There are many solutions to this problem, and perhaps I chose the worst one possible. I hope that someone could come up with a better answer.
This problem is adapted from a recent competition (which is different than all of the "recent competitions" that I have mentioned before.

After getting $(x^2-x+1)\mid(3x-1)$, and bounding $-2\leq x\leq 3$ as a result, you should first go back and check the condition $(x^2-x+1)\mid(3x-1)$:


*

*$x=-2$: $x^2-x+1=7$, $3x-1=-7\quad\checkmark$

*$x=-1$: $x^2-x+1=3$ so doesn't divide $3x-1$.

*$x=0,1$: $x^2-x+1=1\quad\checkmark$

*$x=2$: $x^2-x+1=3$, so doesn't divide $3x-1$.

*$x=3$: $x^2-x+1=7$ does not divide $3x-1=8$.


then find $y$.
