What is the smallest positive integer $n$ such that all the roots of $z^4 - z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity?

The answer isn't 6 @Steve...

To find the smallest positive integer $n$ such that all the roots of $z^4 - z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity, we can start by factoring the given equation.

$$z^4 - z^2 + 1 = 0$$

Let's notice that this equation resembles a quadratic equation in terms of $z^2$. So, we can make a substitution: Let $x = z^2$.

Substituting $x$ into the equation, we get:

$$x^2 - x + 1 = 0$$

Now, we have a quadratic equation. To solve it, we can use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

For our equation, $a = 1$, $b = -1$, and $c = 1$. Plugging in these values, we have:

$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}$$

Simplifying further, we get:

$$x = \frac{1 \pm \sqrt{-3}}{2}$$

Since we are looking for the square root of $x$, we can rewrite this as:

$$x = \frac{1}{2} \left(1 \pm \sqrt{-3} \right)$$

Since $-3$ is not a perfect square, we have complex roots. To find the $n^{\text{th}}$ root of unity, we need the imaginary part of the complex roots to be equal to zero.

Now, let's consider the imaginary part:

$$\text{Im}\left(\frac{1}{2} \left(1 \pm \sqrt{-3} \right)\right) = \frac{\pm \sqrt{3}}{2}$$

To make the imaginary part equal to zero, we need to choose the minus sign:

$$\frac{-\sqrt{3}}{2} = 0$$

Solving this equation, we find:

$$\sqrt{3} = 0$$

This equation is not satisfied for any real number. Therefore, we conclude that there is no smallest positive integer $n$ for which all the roots of $z^4 - z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity.

Hence, the answer is that there is no such positive integer $n$.