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Harvard-MIT Mathematics Tournament

Algebra Difficulty 4.5 AIME Prove it United States

Problem:
Find all real solutions $(x, y)$ of the system $x^{2}+y=12=y^{2}+x$ .

Solution

Solution:
We have $x^{2}+y = y^{2}+x$ which can be written as $(x-y)(x+y-1)=0$ .

The case $x = y$ yields $x^{2}+x-12=0$ , hence $(x, y) = (3, 3)$ or $(-4, -4)$ .

The case $y = 1-x$ yields $x^{2}+1-x-12 = x^{2}-x-11=0$ which has solutions $x = \frac{1 \pm \sqrt{1+44}}{2} = \frac{1 \pm 3\sqrt{5}}{2}$ . The other two solutions follow.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.