# A proof of theorem about primitive roots [closed]

I have a such equation: $${x}^{n}-1=0$$ I have n complex roots. For example:$${x}^{7}-1=0$$ $${x}_{1}=1; {x}_{2} = {(-1)}^{\frac{1}{7}}; {x}_{3}=-{(-1)}^{\frac{2}{7}}; ...;{x}_{7}={(-1)}^{\frac{6}{7}}$$ Then I could say, that one of this roots is primitive. Obviously, it's $${x}_{2} = {(-1)}^{\frac{1}{7}},$$ because all powers of this root enumarate rest of roots. So, I care about theorem, that roots $${x_2}^{0};{x_2}^{1};{x_2}^{2};{x_2}^{3};{x_2}^{4};{x_2}^{5};{x_2}^{6}$$ do not equal between themself. Below I am going to show a proof of the statement:

Let a is primitive root. Let $${a}^{k}={a}^{m}, k \neq m$$ Then $${a}^{k-m}=1,$$but $$(k-m)and, therefore cannot be devided on n . Consequently$${a}^{k}\neq{a}^{m}.$$ Do you find it plain, that if you raise a number to a power, you cannot get two numbers with similar powers? Why that proof is established on the such suppose? It's obvious that I cannot raise in such power the number $${x}_{2}$$ that I obtain, for example, that:$${{x}_{2}}^{2}={(-1)}^{\frac{2}{7}}$$ or that: $${{x}_{2}}^{4}={(-1)}^{\frac{4}{7}}$$ and that they will be equal. Despite my desire, it's impossible to get two equal numbers that have different powers, if I use powers less than n. How did it build the proof?

## closed as unclear what you're asking by mrtaurho, hardmath, Leucippus, Lord Shark the Unknown, ancientmathematicianFeb 8 at 7:39

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• It might help to do a small example, say $n=4$ or $n=6$ to understand that while you can get some powers of the roots to be equal below $n$, not all of the roots will have the same power. A primitive root $r$ will be one that has all its powers distinct up through $r^n = 1$. – hardmath Feb 7 at 14:58

I feel like Adwin's suggestion of using Polar coordinates may be the best'' approach here, but there is an alternative method: In general, a polynomial has a repeated root if and only if it has a common factor with its derivative (this is an easy exercise). Then observe that $$x^n-1$$ is relatively prime to its derivative, $$nx^n$$, so $$x^n-1$$ has no repeated roots.
A quick look suggests that $$(-1)^{\frac{1}{7}}$$ is not a root of the equation. Neither is $$x_3$$.