4. [6] Find the real solution(s) to the equation $(x+y)^{2}=(x+1)(y-1)$.
Answer: $(-1,1)$ Set $p=x+1$ and $q=y-1$, then we get $(p+q)^{2}=p q$, which simplifies to $p^{2}+p q+q^{2}=0$. Then we have $\left(p+\frac{q}{2}\right)^{2}+\frac{3 q^{2}}{4}$, and so $p=q=0$. Thus $(x, y)=(-1,1)$.