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Let $p$ be prime, $p \equiv 1 \pmod{3}$. I wish to show that there are $(p-1)/3$ (non-zero) cubic residues (mod $p$), for which I am having some difficulty.

There is a response here by userabc, which I am not certain how to follow. I can easily prove that $-3 \in Q_{p}$ and show that $x^3 - 1 \equiv 0 \pmod{p}$ admits three solutions. (In fact, if $-3 \equiv c^2$, then these three solutions are given by $1, 2^{-1}(c-1)$ and $-2^{-1}(c-1)$). How next am I to make use of the identity $4(x^2 + x + 1) = (2x+1)^2+3$? And how do I conclude from here that there are $(p-1)/3$ cubic residues?

V.E.
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    If you are allowed to use some group theory, the problem becomes much easier. Define a group homomorphism $\varphi:\mathbb{F_p}^{\times}\to\mathbb{F_p}^{\times}$ by $x\to x^3$. The first isomorphism theorem gives an easy way to compute the cardinality of the image. – Mark Mar 17 '24 at 01:34

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When $x, y \not\equiv 0 \bmod p$, we have $x^3 \equiv y^3 \bmod p$ if and only if $(x/y)^3 \equiv 1 \bmod p$, so $x \equiv yz \bmod p$ where $z^3 \equiv 1 \bmod p$. Thus the cubing function on nonzero numbers mod $p$ is $k$-to-$1$ where $k$ is the number of solutions to $z^3 \equiv 1 \bmod p$. This is similar to squaring on nonzero numbers mod $p$ being $2$-to-$1$ when $p > 2$ since $x^2 \equiv y^2 \bmod p$ if and only if $x \equiv \pm y \bmod p$, where $\pm 1 \bmod p$ are the solutions to $z^2 \equiv 1 \bmod p$, and there are two such $z$ when $p$ is an odd prime.

So you need to show there are three solutions to $z^3 \equiv 1 \bmod p$ when $p \equiv 1 \bmod 3$. Since $$ z^3 - 1 = (z-1)(z^2+z+1), $$ the solutions to $z^3 \equiv 1 \bmod p$, rewritten as $z^3 - 1 \equiv 0 \bmod p$, are $1 \bmod p$ and the solutions to $z^2 + z + 1 \equiv 0 \bmod p$ (which doesn't have $1 \bmod p$ as a solution since $p > 3$). The quadratic formula is valid mod $p$ when $p > 2$: there are two roots to $z^2 + z + 1 \equiv 0 \bmod p$ if and only if the discriminant $1^2 - 4(1) = -3$ is a nonzero square mod $p$.

It remains to show when $p \equiv 1 \bmod 3$ that $z^2 + z + 1 \equiv 0 \bmod p$ has two roots, which would be equivalent to $-3$ is a square mod $p$. Do you know that the nonzero numbers mod $p$ have a primitive root when $p$ is prime or do you know quadratic reciprocity?

KCd
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  • That is, if I am able to argue that there are three solutions to $z^3 \equiv 1 \pmod{p}$, then, I can use the fact that cubing is $3$-to-$1$, to conclude that there must be $(p-1)/3$ possible cubic residues in $\mathbb{Z}_{p}$? – V.E. Mar 17 '24 at 01:56
  • If you are unsure about this, then why do you accept that there are $(p-1)/2$ quadratic residues among the nonzero integers mod $p$ when$p \not= 2$? – KCd Mar 17 '24 at 12:28