Need help with solving the expression $(x + y)^2 = (x + 1)(y - 1)$ for $x$ and $y$ Solve for $x$ and $y$ in
$(x+y)^2=(x+1)(y-1)$. I've figured out that
$$x^2+2xy+y^2=xy-x+y-1$$ but I'm not sure how to proceed. I'm somewhat familiar with solving quadratic equations by completing the square, could that help somehow? I'd appreciate any tips. 
Thanks -Kurns

The only solution is $x=-1,y=1$.
If $x+1,y-1$ are both positive, then AM-GM implies their product is at most the square of their AM which is $\frac{1}{4}(x+y)^2<(x+y)^2$. If they are both negative, then $-x-1,-y+1$ are both positive and so $(x+1)(y-1)=(-x-1)(-y+1)<(-x-y)^2=(x+y)^2$. If one of $x+1,y-1$ is positive and the other negative, then their product is negative. In none of these cases can we have equality.
If one of $x+1,y-1$ is zero, then $x+y$ must be zero and hence both $x+1,y-1$ must be zero.
