HMMT November

Algebra Difficulty 4.9 AIME Find the answer

Find all triples of positive integers $(x, y, z)$ such that $x^{2}+y-z=100$ and $x+y^{2}-z=124$ .

Your answer

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

Solution

Cancel $z$ to get $24=(y-x)(y+x-1)$ . Since $x, y$ are positive, we have $y+x-1 \geq 1+1-1>0$ , so $0<y-x<y+x-1$ . But $y-x$ and $y+x-1$ have opposite parity, so $(y-x, y+x-1) \in\{(1,24),(3,8)\}$ yields $(y, x) \in\{(13,12),(6,3)\}$ . Finally, $0<z=x^{2}+y-100$ forces $(x, y, z)=(12,13,57)$ .

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