Solve:
$2520=(x+y+xy)^2+2xy+2y-3x$
Please help, I tried Solving SFFT and Quadratic, but couldn't end up with a result.
Edit: I solved this much.I factorized it
$((x+1)(y+1)-1)^2+2((x+1)(y+1)-1)-5x$
Then substituted $t=((x+1)(y+1)-1)$
so $t^2+2t-5x=2520$
and after this i am stuck
Edit 2: Solve for Integers
Best Answer
$$2520=x^2+y^2+(xy)^2+2x^2y+2y^2x+2xy+2xy+2y-3x$$ so, if we group in terms of $y$ $$2520=y^2(x^2+2x+1)+y(2x^2+4x+2)+x^2-3x$$ so $$0=y^2\cdot(x+1)^2+2y(x+1)^2+x^2-3x-2520$$ so solving the quadratic equation in the usual way:
$$\Delta=4(x+1)^4-4(x+1)^2(x^2-3x)=4(x+1)^2(x^2+2x+1-x^2+3x+2520)$$ so we have found out that $$\Delta=(2x+2)^2\cdot(5x+2521)$$
but we know
$$y=\frac{-2(x+1)^2\pm\sqrt{\Delta}}{2(x+1)^2}=\pm\frac{\sqrt{\Delta}}{2(x+1)^2}-1=\pm\frac{2(x+1)\sqrt{5x+2521}}{2(x+1)^2}-1=\pm\frac{\sqrt{5x+2521}}{(x+1)}-1$$ so for $y$ to be an integer, $x+1$ must divide $\sqrt{5x+2521}$ and $5x+2521$ must be a perfect square. Let $5x+2521=k^2$ so $x=\frac{k^2-2521}{5}$ so $$\big(\frac{k^2-2521}{5}+1\big)^2\text{ divides } k$$ so $$(k^2-2516)^2\text{ divides } 25k$$ so $$(k^2-2516)^2\leq 25|k|$$ which gives us $k=\pm50,\pm51$. If $k=\pm 50$, then because $x=\frac{k^2-2521}{5}$, x wouldn't be an integer so contradiction. So $k=51$, so $x=\frac{51^2-2521}{5}=16$
Because $x=16$ wecan now see that because $y=\pm\frac{\sqrt{5x+2521}}{(x+1)}-1=\pm3-1$ so $y$ is $2$ or $-4$.
So the solutions are $(x,y)=(16,-4)$ and $(16,2)$