Like Modulo, Only For pi

In order for $R(f(x),x)=0, \frac{f(x)}{x}$ has to be an integer.

We know that $f(x)>0$ for $x \in \mathbb{R}$ , and since $x$ will be positive in our domain, $\frac{f(x)}{x}$ will be positive over our entire domain.

If we define $g(x)=\frac{f(x)}{x}$ , we get:

$g(x)=x-8+\frac{17}{x}$

Looking at $g(x)$ , we can tell that it is continuous over our interval, and will decrease to some minimum value and then increase from there.

Plugging in some values for $g(x)$ , we get

$g(1)=10\\ g(4)=\frac{1}{4}\\ g(2015)=2007+\frac{17}{2015}$

From all of the above information, we know that over the interval $1 \leq x \leq 2015$ , $g(x)$ will decrease from $10$ to some value $0<g(a)<1$ , thus passing through all the integers between $10$ and $1$ . It will also increase from some value $0<g(a)<1$ to $g(2015)=2007+\frac{17}{2015}$ , thus passing through all the integers from $1$ to $2007$ .

This means $g(x)$ will pass through $10+2007=2017$ integers total, and thus our answer is:

$\boxed{2017}$

Entertaining problem!

We want $\frac{x^2-8x+17}{x}$ to be an integer, or, equivalently, we want $f(x)=x+\frac{17}{x}$ to be an integer. The function $f(x)$ is decreasing for $1\leq x\leq \sqrt{17}$ and then increasing, with $f(\sqrt{17})\approx 8.25$ . The function attains the 10 integer values $f(1)=18$ to 9 on the way down and then the 2007 integer values 9 to 2015 on the way up, for a total of $\boxed{2017}$ .

We want $x^2 - 8x + 17x = nx$ for some positive integer $n$ The roots of $x^2 - (n+8)x + 17 = 0$ are $x_\pm \; =\; \tfrac12\left(n+8 \pm \sqrt{(n+8)^2 - 68}\right) \;,$ Elementary algebra shows that $x_- \ge 1$ provided that $n \le 10$ , and that $x_+ \le 2015$ provided that $n \le 2017$ . Thus there are $10+2007 = 2017$ solutions to the problem.