Floor function equation

Instead of focusing on $x$ , let's focus on $n := \lfloor{x}\rfloor$ and $\varepsilon := \{x\}$ . Note that $n$ can be any nonzero integer and $\varepsilon \in [0,1)$ , and both of them can be chosen independently of one another. In these new terms, we have $(n+\varepsilon)n = y \Rightarrow \frac{y}{n} = n + \varepsilon \in [n,n+1).$

Now, if $n > 0$ , this implies $y \in \left[n^2, n^2+n\right) = \left[n^2, n^2+|n|\right)$ , and if $n<0$ , this implies $y \in \left(n^2+n, n^2\right] = \left(n^2-|n|, n^2\right]$ . Combining these, $y$ is an integer such that there is a positive integer $m = |n|$ such that, $y \in \left(m^2-m, m^2+m\right)$

More explicitly, $y \in \bigcup_{m\in \mathbb{N}^+} \left(m^2-m, m^2+m\right) = (0,2) \cup (2,6) \cup (6,12) \cup \cdots$ where the observation that the right endpoint of one open interval and the left endpoint of the next open are the same can be confirmed by verifying $m^2 + m = (m+1)^2 - (m+1)$ for all $m \in \mathbb{N}^+$ .

Finally, we have the way to find our answer: The only positive integers which cannot be the $y$ -value are those integers of the form $m^2+m$ for some $m\in \mathbb{N}^+$ . Solving $1 \leq m^2 + m \leq 2017 \quad \Rightarrow \quad 1 \leq m \leq 44$ so $y$ cannot be $44$ numbers out of the $2017$ positive integers in that range, so there are $2017 - 44 = \boxed{1973}$ possible such values for $y$ .

I first carefully examined the graph and include a snapshot of the bottom:

It became clear that if y was a perfect square number, then there were TWO solutions for x .

If y was of the form $n^2+n$ , then there was NO SOLUTION for x .

If y was any other number, then there was a SINGLE solution for x , sometimes + , sometimes - .

In the range of values for y, from 1 to 2017 inclusive, we find 44 perfect squares.

$2017-44= \boxed{1973}$

This solution is not as rigorous or detailed as @Brian Moehring provides (which I upvoted - thanks, Brian) but it provides a visual way of seeing the problem.