Is it possible for a square root function,f(x), to map to a finite number of integers for all x in domain of f? Consider the equation $$f(x) =\sqrt{x^2 - x + 1}$$
Using python I checked x for $$ -100000000 \leq x \leq  100000000$$
and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?
Edit: $$x \in \mathbb{Z}$$
 A: First of all, observe that the function is defined $\forall x\in \mathbb Z$ since $x^2+1\geq2x\geq x\iff x^2-x+1\geq 0$. 
Completing the square, we get $$x^2-x+1=(x-1)^2+\color{blue}x$$
It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$. 
Furthermore 
\begin{align*}(x-1)^2-(x-2)^2&=\color{blue}{2x-3}\tag{1}\\
x^2-(x-1)^2&=\color{blue}{2x-1}\tag{2}
\end{align*}
Can you end it now?

 Hint: Observe, for instance, that $$\lvert 2x-3\rvert>\lvert x\rvert \text{ unless } x\in[1, 3]$$ $$\lvert 2x-1\rvert>\lvert x\rvert \text{ unless } x\in[\frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $x\in\{1, 2, 3\}$.

A: Hint: let $y = \sqrt{x^2-x+1}$. Squaring both sides,
$$y^2 = x^2-x+1,$$
so $y^2-1=x^2-x$. That is,
$$(y+1)(y-1) = x(x-1).$$
So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?
A: Hint: 
For $x>1$, $(x-1)^2 \lt x^2-x+1 \lt x^2$;  
for $x<0$, $x^2<x^2-x+1<(x-1)^2$.  
