Solve $\lfloor x^2 + 2x \rfloor = \lfloor x^2 \rfloor + 2 \lfloor x \rfloor$

Question

How do we find all real $x$ such that $\lfloor x^2 + 2x \rfloor = \lfloor x^2 \rfloor + 2 \lfloor x \rfloor$, where $\lfloor \space \rfloor$ denotes the "greatest integer function" ?

Answer 1

Note that $$ \newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} \floor{a_1 + a_2 + \cdots + a_k} \ge \floor{a_1} + \floor{a_2} + \cdots + \floor{a_k} $$ with equality if and only if $$ \{a_1\} + \{a_2\} + \cdots + \{a_k\}< 1 $$ where $\{ \space \}$ denotes the fractional part. In particular, $$ \floor{x^2 + 2x} \ge \floor{x^2} + \floor{x} + \floor{x} $$ So if $\{x\} = r < 1, \{x^2\} = s < 1$, you are looking for when $$ s + 2r < 1 $$ There are uncountably many solutions to this. To generate a solution, take any irrational $r$ with $0 < r < \frac12$. Then find any integer $n$ such that $2nr + r^2$ has fractional part less than $1 - 2r$, and let $x = n + r$.