Say p $> 7$ be a prime such that p $\equiv$ 1 (3). Show that the set of cubic residues (including zero) do not form an arithmetic progression.

Question

I already know that $\mathbb{Z}/p\mathbb{Z}$ - $\{0\}$ is partitioned into C$_{p}$, $x C_{p}$ and $ x^{2}C_{p}$ , where 'x' is a non-cubic-residue. This means that the cardinality of set of cubic residues including zero is 'k+1' where p = 3k+1.

Can anyone help me show that residues do not actually form an arithmetic progression? I am totally stumped here.

Answer 1

Hint:

As you indicated, there are $k+1$ cubic residues, including $0$.

$1$ and $p-1$ are cubic residues ($1^3$ and $(-1)^3$, respectively).