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HMMT February

Algebra Difficulty 4.7 AIME Find the answer

Find all real solutions (x,y)(x, y) of the system x2+y=12=y2+xx^{2}+y=12=y^{2}+x.

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

Solution

We have x2+y=y2+xx^{2}+y=y^{2}+x which can be written as (x−y)(x+y−1)=0(x-y)(x+y-1)=0. The case x=yx=y yields x2+x−12=0x^{2}+x-12=0, hence (x,y)=(3,3)(x, y)=(3,3) or (−4,−4)(-4,-4). The case y=1−xy=1-x yields x2+1−x−12=x2−x−11=0x^{2}+1-x-12=x^{2}-x-11=0 which has solutions x=1±1+442=1±352x=\frac{1 \pm \sqrt{1+44}}{2}=\frac{1 \pm 3 \sqrt{5}}{2}. The other two solutions follow.

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