# 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: $$x \in \mathbb{Z}$$

• To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers. – Théophile Apr 5 at 20:08
• Well, if $x^2-x-3=0$ (that is, if $x=(1+\sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$. – Arturo Magidin Apr 5 at 20:11
• Justed edited, yes I meant to say $$x \in \mathbb{Z}$$ – Diehardwalnut Apr 5 at 21:28

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 $$x\in\{1, 2, 3\}$$.

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)?

Hint:

For $$x>1$$, $$(x-1)^2 \lt x^2-x+1 \lt x^2$$;

for $$x<0$$, $$x^2.