Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory.

How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$?

Really, even pointers on how to say anything meaningful about these $p$ are welcome. Originally I also asked about how to count the density of all $p$ (not just those of the form $3k+1$) such that $2$ (or $3$) is not a cubic residue modulo $p$, but Felipe Voloch's comment quickly addresses how to deal with them, via Chebotarev's density theorem.

The difference between the question and these easier problems is that here I am asking that $k+1$ is not a cube modulo the prime $3k+1$, so the same approach does not seem to apply.

Finally, if it turns out that the density is not zero, how does one go about finding the density of those $p=3k+1$ that satisfy that none of the equations $x^3=2$, $x^3=3$, $x^3=k+1$ have solutions?

(Ideally, the techniques lift to other situations, such as studying fifth powers modulo primes $p=5k+1$, etc, but even methods exclusive to the case of cubes are very welcome.)

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    $\begingroup$ You are looking for the primes that don't split completely in the splitting field of $x^3-2$ but split in the subfield of the cubic roots of unity, so you can apply Cebotarev. I think the density is $2/3$ among primes $1 \pmod 3$. $\endgroup$ Sep 12, 2013 at 19:36
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    $\begingroup$ Your conditions are equivalent to saying something about the Frobenius conjugacy class of $p$ in the Galois closure ${\mathbf Q}(\zeta_3,\sqrt[3]{2/3}) = {\mathbf Q}(\zeta_3,\sqrt[3]{18})$: you want $p$ to be split in ${\mathbf Q}(\zeta_3)$ and be inert in ${\mathbf Q}(\sqrt[3]{18})$, which is equiv. to its Frobenius conj. class in ${\rm Gal}({\mathbf Q}(\zeta_3,\sqrt[3]{18})/{\mathbf Q})$ being the 3-cycles, so by Chebotarev the density is $2/6 = 1/3$. $\endgroup$
    – KConrad
    Sep 14, 2013 at 12:38
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    $\begingroup$ Did you do any numerical experiments? If you sample your condition over the first 1000 primes, the proportion fitting your condition is 334, and 334/1000 = .334. Also, insisting that there is no solution to $2 \equiv 3x^3 \bmod p$ forces $p \equiv 1 \bmod 3$, since when $p \equiv 2 \bmod 3$ every number mod $p$ is a cube, as 3 is relatively prime to $p-1$. Thus your constraint that $p \equiv 1 \bmod 3$ is automatic given your other constraint that $2 \equiv 3x^3 \bmod p$ has no solution. $\endgroup$
    – KConrad
    Sep 14, 2013 at 12:49
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    $\begingroup$ In your last question, the condition that $p \equiv 1 \bmod 3$ is automatic as soon as you ask for something mod $p$ not to be a cube. You want to count the proportion of $p$ such that $2$, $3$, and $2/3$ are all not cubes. This can be expressed in terms of condition on the Frobenius conjugacy class of $p$ in ${\rm Gal}({\mathbf Q}(\zeta_3,\sqrt[3]{2},\sqrt[3]{3})/{\mathbf Q})$, so it can be answered by the Chebotarev density theorem too. $\endgroup$
    – KConrad
    Sep 14, 2013 at 12:58
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    $\begingroup$ I ran a numerical experiment on the first 10000 and 20000 primes for your last question. The proportions are around .11, and the answer to your last question is a rational number with denominator 18 (since the Galois group of the splitting field of $x^3-2$ and $x^3-3$ has order 18), so the correct answer is presumably $2/18 = 1/9 = .1111...$, and to prove that amounts to showing the primes $p$ satisfying the conditions in your last question have a common Frobenius conjugacy class of size 2 in the Galois group. $\endgroup$
    – KConrad
    Sep 14, 2013 at 13:13

1 Answer 1


A solution can be obtained as suggested by Keith Conrad in the comments, via Chebotarëv's theorem. Details can be found in $\S3.4$ of

Coloring the $n$-Smooth Numbers with $n$ Colors

Andrés Eduardo Caicedo, Thomas A. C. Chartier, Péter Pál Pach

The Electronic Journal of Combinatorics 28 (1) (2021), #P1.34, 79 pp. DOI: https://doi.org/10.37236/8492

Many thanks to both Felipe Voloch and David E Speyer for their patient explanations.


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