Question 128
Choose any rational point $P_0 = (x_p, y_p)$ on the ellipse $C: x^2 + 2y^2 = 2098$. We determine $P_1, P_2, \cdots$ recursively as follows: for all $i = 0, 1, \cdots,$
(1) choose a lattice point $Q_i = (x_i, y_i)$ not on $C$ such that $|x_i| \lt 50$ and $|y_i| \lt 50$;
(2) draw $\overline{P_iQ_i}$ and let $P_{i+1}$ be its other intersection point with $C$.
Prove that for any $P_0$ we can choose $Q_0, Q_1, \cdots$ suitably so that there is a non-negative integer $k$ with $\overline{OP_k} = \sqrt{2017}$.
(We call $(x, y)$ a lattice point if and only if $x$ and $y$ are both integers. We call $(x, y)$ a rational point if and only if $x$ and $y$ are both rational.)
(1) choose a lattice point $Q_i = (x_i, y_i)$ not on $C$ such that $|x_i| \lt 50$ and $|y_i| \lt 50$;
(2) draw $\overline{P_iQ_i}$ and let $P_{i+1}$ be its other intersection point with $C$.
Prove that for any $P_0$ we can choose $Q_0, Q_1, \cdots$ suitably so that there is a non-negative integer $k$ with $\overline{OP_k} = \sqrt{2017}$.
(We call $(x, y)$ a lattice point if and only if $x$ and $y$ are both integers. We call $(x, y)$ a rational point if and only if $x$ and $y$ are both rational.)