Drop it Like it's a Floor Function

$nx^2=n\lfloor{x^2}\rfloor+x \Rightarrow n(x^2-\lfloor{x^2}\rfloor)=x \Rightarrow n\{x^2\}=x$

where $\{\cdot \}$ represents the fractional part function.

Now substitute $x^2=t$ , so the equation is now

$\{t\}=\frac{\sqrt{t}}{n}$

The question can now be solved with some graphical intuition. Clearly, $y=\sqrt{x}/n$ represents a parabola. This parabola should intersect the curve $y=\{x\}$ $2014$ times as per the question. One point of intersection is origin. For the lower bound of $n$ , the parabola must pass through a point $(2013+\delta,1)$ where $\delta \rightarrow 0^+$ , the corresponding value of $n$ for this is $\sqrt{2013}$ . The maximum value of $n$ is achieved when the parabola passes through $(2014,1)$ . Hence,

$n\in (\sqrt{2013},\sqrt{2014}]$

We have the equation $nx^2=n\lfloor x^2\rfloor +x$

First, we isolate the $x$ , and try to factor some stuff: $n(x^2-\lfloor x^2\rfloor)=x$

Now let's divide by $n$ : $x^2-\lfloor x^2\rfloor=\dfrac{x}{n}$

Thus, we want the number of intersections of the graph $f(x)=x^2-\lfloor x^2\rfloor$ and the linear graph $g(x)=\dfrac{x}{n}$ to be exactly $2014$ .

Let's try to find out what exactly the graph $f(x)=x^2-\lfloor x^2\rfloor$ makes. First, we see that if $x=\sqrt{a}$ for some non-negative integer $a$ , then $f(x)=0$ . Furthermore, we see that if $x=\sqrt{a}-\epsilon$ for some small $\epsilon > 0$ , then we have $x^2\approx a$ and $\lfloor x^2\rfloor \approx a-1$ so $f(x)\approx 1$ .

Thus, the graph's range is $[0,1]$ . How does that help us? Well, if $\dfrac{x}{n} \ge 1$ , then we know that any $x$ equal or greater will not produce any solutions!

Let's see what else we can find out from this graph. First, we see that the growth of the function from $x=0$ to $x=1$ is the same as $x^2$ . Then, from $x=1$ to $x=\sqrt{2}$ , the graph is reset to $0$ , and then exhibits the same grown as $x^2$ from $x=1\to \sqrt{2}$ . Ditto for $x=\sqrt{2}\to \sqrt{3}$ and so on. Thus, the graph is increasing for all intervals $[\sqrt{n},\sqrt{n+1})$ .

Now let's investigate where $x$ will generate a solution.

First off, we see that every time $f(x)=0$ , it generates one solution. This is because the graph is increasing from $f(x)=0$ to $f(x) < 1$ . Thus, every time $x=\sqrt{a}$ for a non-negative integer $a$ , if $\dfrac{x}{n} < 1$ at $x=\sqrt{a}$ , then there will be a solution.

Thus, we want there to be solutions for $x=\sqrt{0}\to \sqrt{2013}$ . This means $\dfrac{\sqrt{2013}}{n} < 1$ so $n > \sqrt{2013}$ . Similarly we want $\dfrac{\sqrt{2014}}{n}\ge 1$ to ensure there isn't $2015$ solutions. Solving gives $n\le \sqrt{2014}$ .

Thus, $n\in (\sqrt{2013},\sqrt{2014}]$ so our answer is $\lfloor 1000\sqrt{2014}-1000\sqrt{2013}\rfloor=\boxed{11}$ .

Here is an interactive graph to help your understanding.