How many positive real numbers $x\le 2000$ exist such that $x\cdot \left\{ x \right\}=17$? I've been looking at the ceiling/floor functions and the fractional part function ($\left\{ x \right\}$) and I came across this type of problem. I am aware of the series expansions one can use the express $\left\lfloor x \right\rfloor$ and therefore $\left\{ x \right\}$, but I don't think that provides a way to deal with this (at least not directly). Is there any approach that allows us to enumerate the number of cases without having to resort to brute calculation? Could this be generalized to any for any $n$ for which $x\le n$ ? I would be very grateful for insights into this.
(Insterestinly, WA provides a single $x\cong 30.55$.I am not sure how plausible this is, especially considering the shape of the graph.)
 A: Let $x=n+\varepsilon$ where $n = \lfloor x \rfloor$ and $\varepsilon = \{x\}$. Then there are solutions for $x \{x\}=17$ if and only if
$$
\varepsilon^2 + n\varepsilon - 17 = 0.
$$
Solve for $\varepsilon$ and find solutions where $\varepsilon\in[0,1)$.
$$
\varepsilon_{1,2}=\frac{-n\pm\sqrt{n^2+4*17}}{2}.
$$
One root is negative and the other is positive, so we must have
$$
\varepsilon=\frac{-n+\sqrt{n^2+4*17}}{2}.
$$
Since $\varepsilon \in [0,1)$ this boils to
$$
\sqrt{n^2+4*17} < 2+n,\\
n^2+4*17 < 4+4n+n^2 .
$$
Therefore we have a solution for any $n > 16$
A: From the equation, $x$ must be greater than $17$. Suppose $x = k + \{x \}$ with $k\in \Bbb N$ and $17 \le k  \le 2000$. Then
$$(k + \{x \})\{x \} = 17$$
This quadratic equation on $\{x \}$ has 1 positive real root
$$\{x \} = \frac{\sqrt{68+k^2}-k}{2}$$
We can verify easily that this root is in $(0,1)$ with $k \ge 17$.
Then, all solution of the equation is
$$x = k +  \frac{\sqrt{68+k^2}-k}{2} =  \frac{\sqrt{68+k^2}+k}{2} \qquad,k = 17,...,1999$$
Then there is $1999-17+1 = 1983$ solutions.
A: Decompose $x$ into integral and fractional parts $n+a$. We have $(n+a)a=17$ or $a^2+na-17=0$ or $a=\frac{-n\pm\sqrt{n^2+68}}2$. Now we cannot take the minus option since $\frac{\sqrt{n^2+68}}2>1$ and $a$ has to lie in $[0,1)$, so $2a=\sqrt{n^2+68}-n$ ($2a\in[0,2)$).
The function $\sqrt{n^2+68}-n$ is monotonically decreasing and conveniently $n=16$ gives exactly $2$, so the solutions $x$ correspond to all integers from $17$ to $1999$ inclusive; they number $1999-17+1=1983$.
