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Problem 994

AMC 12 late, AIME early

Algebra Difficulty 4.9 Find the answer HMMT February

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

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

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

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.