Problem 1802

1. It is easy to see that all the integer points on $C$ are $(\pm44, \pm9)$, and $44^2 + 9^2 = 2017$, so we only need to prove that after appropriate operations, some $P_k$ is an integer point.

2. If $P_0$ is an integer point, then by 1. we may take $k=0$, so assume $P_0 = (a/m, b/m)$ is not an integer point, where $a, b, m \in \mathbb{Z}$ and $m > 0$.

– Clearly there exist integers $s$ and $t$ satisfying $|s - \frac{a}{m}| \le 1/2$ and $|t - \frac{b}{m}| \le 1/2$. Also, $|s| < \sqrt{2098} + 1 < 50$, and similarly $|t| < 50$, so we may take $Q_0 = (s, t)$.

Note that $(s - \frac{a}{m})^2 + 2(t - \frac{b}{m})^2 = 2098 + s^2 + 2t^2 - 2(sa + 2tb)/m = m'/m$, where $m' \in \mathbb{N}$. But by definition, we have
$$\left| \left(s - \frac{a}{m}\right)^2 + 2\left(t - \frac{b}{m}\right)^2 \right| \le 1/4 + 2/4 < 1,$$
so $m' = m \left( \left(s - \frac{a}{m}\right)^2 + 2\left(t - \frac{b}{m}\right)^2 \right) < m$.

– Now, suppose $P_1 = (s+z(a-sm), t+z(b-tm))$, then we have $(s+z(a-sm))^2 + 2(t+z(b-tm))^2 = 2098$, and expanding gives
$$\frac{m'}{m}z^2 + 2(s(a - sm) + 2t(b - tm))z + (s^2 + 2t^2 - 2098) = 0.$$
One solution of the above equation is $z = 1/m$ (corresponding to the solution for $P_0$), so by Vieta's formulas, the other solution is $(s^2 + 2t^2 - 2098)/m'$. Therefore $P_1 = (a'/m', b'/m')$, where $a', b', m' \in \mathbb{Z}$ and $0 < m' < m$.

- From the above discussion, if $Q_i$ is chosen in the above manner at every step, then the denominator of the coordinates of the resulting $P_{i+1}$ will be smaller than the denominator of the coordinates of $P_i$. This means that there exists a sufficiently large $k$ such that $P_k$ is an integer point $\Rightarrow \overline{OP_k} = 2017$. This completes the proof.