Problem 1518

Solution:

The complex number $\omega=\frac{-1+\sqrt{3} i}{2}$ is, as is well known, a cube root of unity and satisfies $\omega^{2}=\frac{-1-\sqrt{3} i}{2}$, $\omega^{3}=1$ and $\omega^{2}+\omega+1=0$; in particular
$$1+\omega^{j}+\omega^{2 j}= \begin{cases}3, & \text{ if } j \text{ is divisible by three } \\ 0 & \text{ otherwise. }\end{cases}$$
Set $r_{k}=\sqrt[3]{28} \omega^{k}-3$ for $k=0,1,2$. By the definition of $b$ we have $\left|r_{0}^{-n}-b\right|<1$; since the real parts of $\omega$ and $\omega^{2}$ are negative, we have $\left|r_{1}\right|>1$ and $\left|r_{2}\right|>1$. Thus
$$\left|b-\left(r_{0}^{-n}+r_{1}^{-n}+r_{2}^{-n}\right)\right|<\left|b-r_{0}^{-n}\right|+\left|r_{1}^{-n}\right|+\left|r_{2}^{-n}\right|<3 .$$
Since $\left(\sqrt[3]{28} \omega^{k}\right)^{3}=28$, we have
$$r_{k}^{-1}=\frac{1}{\sqrt[3]{28} \omega^{k}-3}=\frac{\left(\sqrt[3]{28} \omega^{k}\right)^{3}-3^{3}}{\sqrt[3]{28} \omega^{k}-3}=\sqrt[3]{28}^{2} \omega^{2 k}+3 \sqrt[3]{28} \omega^{k}+9$$
If one raises the polynomial $X^{2}+3 X+9$ to the $n$-th power, there exist integers $c_{0}, \ldots, c_{2 n}$ with $\left(X^{2}+3 X+9\right)^{n}=c_{2 n} X^{2 n}+c_{2 n-1} X^{2 n-1}+\ldots+c_{0}$, where $c_{0}=9^{n}$ is odd. Substituting $X=\sqrt[3]{28} \omega^{k}$ gives $r_{k}^{-n}=\sum_{j=0}^{2 n} c_{j} \sqrt[3]{28}^{j} \omega^{k j}$; from this, using (1), we obtain:
$$r_{0}^{-n}+r_{1}^{-n}+r_{2}^{-n}=\sum_{j=0}^{2 n} c_{j} \sqrt[3]{28}^{j}\left(1+\omega^{j}+\omega^{2 j}\right)=3 \sum_{0 \leq \ell \leq 2 n / 3} c_{3 \ell} 28^{\ell}$$
The sum is clearly a multiple of 3 and, moreover, odd, since the summand $c_{3 \ell} 28^{\ell}$ is odd for $\ell=0$ and even for $\ell>0$. If $b$ were divisible by 6, then the absolute value $\left|b-\left(r_{0}^{-n}+r_{1}^{-n}+r_{2}^{-n}\right)\right|$ would be at least 3, contradicting (2).