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Algebra Difficulty 4.9 AIME Find the answer

Find the real solution(s) to the equation (x+y)2=(x+1)(y−1)(x+y)^{2}=(x+1)(y-1).

A number or a short expression. Spacing and $ signs are ignored.

Solution

Set p=x+1p=x+1 and q=y−1q=y-1, then we get (p+q)2=pq(p+q)^{2}=pq, which simplifies to p2+pq+q2=0p^{2}+pq+q^{2}=0. Then we have (p+q2)2+3q24\left(p+\frac{q}{2}\right)^{2}+\frac{3q^{2}}{4}, and so p=q=0p=q=0. Thus (x,y)=(−1,1)(x, y)=(-1,1).

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.