## Problem 4

Let $k$ be the real cube root of 2. Find integers A, B, C, a, b, c such that $\left|\left(A x^{2}+B x+C\right) /\left(a x^{2}+b x+c\right)-k\right|<\mid$ $\mathrm{x}-\mathrm{k} \mid$ for all non-negative rational $\mathrm{x}$.


## Solution

Taking the limit, we must have $\left(\mathrm{Ak}^{2}+\mathrm{Bk}+\mathrm{C}\right)=\mathrm{k}\left(\mathrm{ak}^{2}+\mathrm{bk}+\mathrm{c}\right)$, so $\mathrm{A}=\mathrm{b}, \mathrm{B}=\mathrm{c}, \mathrm{C}=2 \mathrm{a}$. Now notice that $\left(b x^{2}+c x+2 a\right)-k\left(a x^{2}+b x+c\right)=(b-a k) x^{2}+(c-b k) x+(2 a-c k)=(x-k)\left((b-a k) x+c-a k^{2}\right)$. So we require $\left|(b-a k) x+c-\mathrm{ak}^{2}\right|<\left|a x^{2}+b x+c\right|$ for all $x \geq 0$.

There are many ways to satisfy this. For example, take $\mathrm{a}=1, \mathrm{~b}=\mathrm{c}=2$. Then ( $\mathrm{b}-\mathrm{ak}) \mathrm{x}$ is always positive and less than $\mathrm{bx}$ for positive $\mathrm{x}$, and $\mathrm{c}-\mathrm{ak}^{2}$ is positive and less than $\mathrm{c}$.

