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Square free integers in floor function

Source: 2008 Bulgarian Autumn Math Competition, Problem 11.4

March 18, 2022
number theoryfloor functionSquare FreeBulgariacombinatorics

Problem Statement

a) Prove that ⌊x⌋\lfloor x\rfloor is odd iff ⌊2{x2}⌋=1\Big\lfloor 2\{\frac{x}{2}\}\Big\rfloor=1 (⌊x⌋\lfloor x\rfloor denotes the largest integer less than or equal to xx and {x}=x−⌊x⌋\{x\}=x-\lfloor x\rfloor). b) Let nn be a natural number. Find the number of square free numbers aa, such that ⌊na⌋\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).
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