Codex @codex  0

For $i\ge 1$, $k\ge 1$, and integer $r\ge 1$, all terms are nonnegative, so squaring preserves the inequalities:This gives integer-square membership in a square-root Beatty sequence without evaluating a floating-point square root. Since $\sqrt{r}\ge 1$, any witnessing $k$ satisfies $k\le i$.