From CMC:
What is the sum of the square of the real numbers $x$ for which $x^2 – 20\lfloor x\rfloor + 19 = 0$?
We use $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$ and eventually get the bounds $1\le x\le19$ and $x\ge 18,x\le 2.$ Of course, it's possible for $x$ not to be an integer, so how do we find the other solutions, other than $19$ and $1$?
Someone wrote this solution:
$x^2 – 20\lfloor x \rfloor + 19 = 0$
Cleary $x\geq \lfloor x \rfloor$ for all real $x$. Thus,
$$x^2-20x+19 \leq x^2 – 20\lfloor x \rfloor + 19=0.$$
Which leads to
$$1 \leq x \leq19.$$Also $x^2=20\lfloor x\rfloor – 19$ which implies $\lfloor x \rfloor=1,17,18,19$.
I'm not sure how we get $\lfloor x\rfloor=17,18$ from this.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{x^{2} - 20\left\lfloor\,{x}\,\right\rfloor + 19 = 0}\,,\quad x = {\Large ?}}$
It is clear that $\ds{\left\lfloor x\right\rfloor = {x^{2} + 19 \over 20} = m, \mbox{where}\ m \in \mathbb{N}_{\geq\ 1}\ \mbox{such that}\ x = \root{20m - 19}}$.
Then, \begin{align} &\bbox[5px,#ffd]{m = \left\lfloor\,{\root{20m - 19}}\,\right\rfloor} \implies m \leq \root{20m - 19} < m + 1 \\[5mm] & \implies m^{2} \leq 20m - 19 < m^{2} + 2m + 1 \implies \left\{\begin{array}{lcl} \ds{m^{2} - 20m + 19} & \ds{\leq} & \ds{0} \\ \ds{m^{2} - 18m + 20} & \ds{>} & \ds{0} \end{array}\right. \\[5mm] &\ \mbox{with solutions}\quad 1 \leq m <\ \underbrace{9 - \root{61}}_{\ds{\approx 1.1898}}\ \quad\mbox{or}\quad \underbrace{9 + \root{61}}_{\ds{\approx 16.8102}}\ < m \leq 19 \\[5mm] &\ \implies m \in \braces{1,17,18,19} \implies \bbx{x \in \braces{1,\root{321},\root{341},19}} \\ & \end{align} with $\ds{\root{321} \approx 17.9165}$ and $\ds{\root{341} \approx 18.4662}$. Please, check for $\ds{\color{red}{x < 0}}$.