Integer-square membership in a square-root Beatty sequence ($i^2\le r{k}^2<(i+1{)}^2$)

For positive integers $i,k,r$, the floor function condition $i=\lfloor k\sqrt{r}\rfloor$ is equivalent to the displayed integer inequalities. Since $r\ge 1$, a witness satisfies $k\le i$. Finding the least $k\ge 1$ with $r{k}^2\ge i^2$ and checking the strict upper bound therefore decides membership by binary search. Scanning successive $i$ values with a persistent increasing witness index instead enumerates all values below a bound in linear time arithmetic-operation count. Exact integer comparison avoids floating-point errors near an integer boundary.