When is 2 a cubic residue mod p?

Question

It's equivalent to compute the primes $p$ such that 2 is not a cubic residue mod $p$. Let $p$ be a prime, now the the map $ \rho: \mathbb{F}_p^\times \rightarrow \mathbb{F}_p^\times $ given by $\rho(n)=n^3$, is not a bijection if and only if it has non-trivial kernel, this only happens if $x^2+x+1=0$ has a solution mod $p$. But this is possible only if $ \left( \frac{-3}{p} \right) =1 \Leftrightarrow p= 1 $ mod $3$.

I started computing the primes under $300$ in which $2$ is not a cubic residue and these are: $$7, 13, 19, 37, 61, 67, 73, 79, 97, 103, 139, 151, 163, 181, 193, 199, 211, 241, 271.$$ Approximately $2/3$ of the primes have the desired property, hence I was thinking we needed to use congruences mod $3^n$, for some $n$, but I'm stuck. Does anyone know how to find the answer?

The Galois group of $x^3-2$ over $\Bbb{Q}$ is not abelian, so the set of primes $p$ such that $2$ is a cubic residue cannot be described by congruences alone. Says Class Field Theory. — Jyrki Lahtonen, Dec 18 '20 at 15:37
IIRC it is possible to list binary quadratic forms with the property that $2$ is a cubic residue modulo $p$ if and only if $p$ is the value of one of those forms. I am somewhat optimistic about this particular having been covered on the site already. Otherwise, wait for Will Jagy (or some other knowledgable user) to show up. — Jyrki Lahtonen, Dec 18 '20 at 15:39
That Galois group is isomorphic to $S_3$. Two thirds of its elements have a fixed point. So, by Chebotarev, for two thirds of the primes the congruence $x^3\equiv2\pmod p$ has a solution. — Jyrki Lahtonen, Dec 18 '20 at 15:42
You can expect answers like this. I suspect that this will be a bit simpler. — Jyrki Lahtonen, Dec 18 '20 at 15:56
A fixed point as in the 2-cycle $\alpha=(13)\in S_3$ has $2$ as a fixed point, $\alpha(2)=2$. — Jyrki Lahtonen, Dec 18 '20 at 17:00
@Jyrki theorem of Gauss, reference http://zakuski.utsa.edu/~jagy/Hudson_Williams_1991.pdf — Will Jagy, Dec 18 '20 at 17:49
And how could I show that $2/3$ of the elements are fixed then? — Alessandro, Dec 18 '20 at 18:47
List the elements of $S_3$: $(1)$, $(12)$, $(13)$, $(23)$, $(123)$, $(132)$. The first four have a fixed point, the last two don't. That's four out of six, or $2/3$. — Jyrki Lahtonen, Dec 18 '20 at 22:46

score 4 · Answer 1 · answered Dec 18 '20 at 17:47

4

The primes that are $1 \pmod 3$ for which $2$ is a cubic residue are all of form $$ p= x^2 + 27 y^2 $$ with integers $x,y$

The primes that are $1 \pmod 3$ for which $2$ is not a cubic residue are all of form $$ q= 4u^2 + 2uv + 7 v^2 $$ with integers $u,v,$ and we allow them negative or positive as needed

answered Dec 18 '20 at 17:47

Will Jagy

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Are you using ideal decomposition in some field such as $\mathbb{Q}[3^{\frac{1}{3}}]$ to get this result? – Alessandro Dec 18 '20 at 18:33
https://en.wikipedia.org/wiki/Cubic_reciprocity – Will Jagy Dec 18 '20 at 19:14
are there any basic proofs? without using these Algebraic tools? – Saravanan Mar 10 '21 at 07:54

When is 2 a cubic residue mod p?

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