Root of unity

$\underset{n\to \infty }{\text{lim}}$ $\left\| n+\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) (n-1)+\left(-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) (n-2)\right\| \Longrightarrow \sqrt3$

There are lots of ways of thinking about this problem. I'll use a couple in the solution below, but there are others.

Claim: the minimum value of $|a+b\omega+c\omega^2|$ across all triples of distinct integers $a,b,c$ is $\sqrt3$ .

Step 1: $\sqrt3$ is achievable. The triple $(a,b,c)=(-1,0,1)$ works

Step 2: we can't do better than $\sqrt3$ .

Because it's easier to work with, define $F(a,b,c)=|a+b\omega+c\omega^2|^2$

We want to show that $F(a,b,c)\ge 3$ for all triples of distinct integers $a,b,c)$ . First, expanding the right-hand side gives

$\begin{aligned} F(a,b,c)&=a^2+b^2+c^2-bc-ca-ab \end{aligned}$

We see that $F$ must be an integer. Since $F$ is certainly non-negative, it suffices to show $F\neq 2$ , $F\neq 1$ and $F\neq 0$ case by case.

Step 2a: $F\neq 2$ (number theory)

We have $F=(a+b+c)^2-3(bc+ca+ab)$ . But all square numbers are congruent to $0$ or $1$ modulo $3$ ; so $F\neq 2$ .

Step 2b: $F\neq 1$ (complex geometry)

Now note that $\omega^2=-1-\omega$ , so we have $F=|(a-c)+(b-c)\omega|^2$ . $1$ and $\omega$ can be thought of as a basis for the complex plane (that is, every complex number $z$ can be written in a unique way in the form $u+\omega v$ where $u$ and $v$ are real). The points with integer coordinates in this basis form a lattice of rhombi.

For a solution with $F=1$ , we need the point given by $u=a-c$ and $v=b-c$ to be on the unit circle centred at $0$ . Since $(a,b,c)$ are distinct, we have

$\begin{aligned} a-c &\neq0 \\ b-c &\neq0 \\ a-c &\neq b-c \end{aligned}$

or in terms of $u,v$ ,

$\begin{aligned} u &\neq0 \\ v &\neq0 \\ u &\neq v \end{aligned}$

However, it can quickly be checked that none of the six lattice points that lie on the unit circle (which form the vertices of a regular hexagon) satisfy all of these conditions.

Step 2c: $F\neq 0$ (algebra)

$\begin{aligned} \omega&=\frac{-1+i\sqrt3}{2} \\ F&=\left| a-\frac{b+c}{2}+i\frac{b-c}{2} \sqrt3 \right|^2 \end{aligned}$

If $F=0$ , then $a-\frac{b+c}{2}+i\frac{b-c}{2} \sqrt3=0$ . But this can only happen if $a-\frac{b+c}{2}=0$ and $\frac{b-c}{2}=0$ , which only happens when $a=b=c$ ; but this is a contradiction, as $a,b,c$ are distinct.

Hence we have shown that $F\ge3$ , and so the minimum value of $F$ is $3$ , and the answer is $\boxed{\sqrt3}$