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How can I solve for $x$ given $a=x \lfloor x \rfloor$

Where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, and where $a$ is a rational number.

What I've done

\begin{align} \frac{a}{x} &= \lfloor x \rfloor \\ \implies \frac{a}{x} & \le x < \frac{a}{x}+1 \\ \end{align}

which yields two cases

\begin{align} \sqrt{a} &\le x \quad (1) \\\\ \left(x+\frac{1+\sqrt{1+4a}}{2}\right) \left(x+\frac{1-\sqrt{1+4a}}{2}\right) &< 0 \quad (2) \end{align}

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  • $\begingroup$ What have you done so far? $\endgroup$ – saulspatz Mar 30 '18 at 3:44
  • $\begingroup$ Compare the graphs of $x\cdot \lfloor x\rfloor$ to $x^2$. You should be able to see that there are values of $a$ for which no such $x$ exists. You should also be able to see that by knowing the value of $a$, you should be able to figure out the value of $\lfloor x\rfloor$, and from there you will be able to calculate $x$ itself. $\endgroup$ – JMoravitz Mar 30 '18 at 3:53
  • $\begingroup$ I know that I could solve it like that, but I wanted to know if it could be solved analytically. $\endgroup$ – Alex Hal Mar 30 '18 at 3:55
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If $\lfloor x \rfloor = n$, you want $x = a/n$ and $n \le a/n < n+1$. Thus (assuming $n > 0$) $n^2 \le a < n^2 + n$. Now $(n+1)^2 = n^2 + 2 n + 1 > n^2 + n$. So:

Given $a \ge 1$, take $n = \lfloor \sqrt{a} \rfloor$. If $a \ge n^2 + n$ there is no solution. Otherwise, $x = a/n$.

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