# $p$ is odd prime such that $p\equiv 1\mod{3}$ and a is primitive root $\mod{p}$

prove or disprove:

Suppose that p is odd prime such that $p\equiv 1\mod{3}$ and a is primitive root mod $p$. Let $a=r^{\frac{p-1}{3}}$ then $1$, $a$, $a^2$ are solution to $x^3\equiv 1\mod{p}$ that are distinct $\mod{p}$.

I guess it is true statement

since $p$ is odd prime such that $p\equiv 1\mod{3}$ then $p-1+3k$ also, since a is primitive root $\mod{p}$ then $(a,p)=1$

Let $a=r^{\frac{p-1}{3}}$ so let $a^3=r^{p-1}$

any help with that please how can I conclude that $1$, $a$, $a^2$ are solution to $x^3\equiv1\mod{p}$ that are distinct $\mod{p}$?

• What did you try? What is $r$? Dec 13, 2017 at 19:18
• Do you know that the multiplicative group of nonzero residue classes modulo $p$ is always cyclic? Dec 13, 2017 at 19:23
• @DietrichBurde r is $r^$a_i$=$n_i where $n_i$ from 1 to p-1 Dec 13, 2017 at 19:37

If $r$ is a primitive root, then $a$ has order $3$ and so generates a subgroup of order $3$. The elements of this subgroup are exactly $1,a,a^2$ and so these are the roots of $x^3=1$. There can be no others because the integers mod $p$ form a field.
• @dr.rise, $a$ has order $3$.