a) Prove that $\lfloor x\rfloor$ is odd iff $\Big\lfloor 2\{\frac{x}{2}\}\Big\rfloor=1$ ($\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$ and $\{x\}=x-\lfloor x\rfloor$). b) Let $n$ be a natural number. Find the number of square free numbers $a$, such that $\Big\lfloor\frac{n}{\sqrt{a}}\Big\rfloor$ is odd. (A natural number is square free if it's not divisible by any square of a prime number).
Problem
Source: 2008 Bulgarian Autumn Math Competition, Problem 11.4
Tags: number theory, floor function, Square Free, Bulgaria, combinatorics