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Are the cube roots of two integers chosen from a uniform distribution between $1$ and $p-1$ inclusive, $p$ prime, essentially evenly distributed? Note that I will use a $p$ such that $p$ is not equivalent to $2$ modulo $3$.

In other words, I'm trying to ensure that if I pick a random number between $1$ and $p-1$ I will have an equal chance of that number being the cube of some integer.

I've searched the site for answers, but my search did not come up with anything, and I don't know enough theory, except for maybe trying to read more on cubic reciprocity.


I've found a similar result for square roots here.

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    $\begingroup$ If $p\equiv1\pmod3$ then there are $\frac13(p-1)$ cubic residues modulo $p$. Does that answer your question? $\endgroup$ – Angina Seng Dec 2 '19 at 4:26
  • $\begingroup$ $a^3\equiv b $ doesn't have solutions for all $b\pmod p$ $\endgroup$ – J. W. Tanner Dec 2 '19 at 4:28
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    $\begingroup$ @LordSharktheUnknown: I think I realized why my question doesn't make sense... I pick two numbers, $\sqrt[3]{x}$ and $\sqrt[3]{y}$. Then I go through a bunch of primes trying to determine which of these cube roots will "exist" modulo the primes. I'm hoping that I can treat them as "evenly distributed", or having equal probability of "existing". $\endgroup$ – Matt Groff Dec 2 '19 at 4:29
  • $\begingroup$ @J.W.Tanner: See my reply to LordSharktheUnknown... I'm really just trying to find when the roots exist modulo various $p$, $p$ prime. $\endgroup$ – Matt Groff Dec 2 '19 at 4:31
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    $\begingroup$ Assuming that $p$ is a large prime $\equiv1\pmod3$ then the so called Polya-Vinogradov method for studying incomplete character sums leads to the result that an interval $[a,b]$, $0<a<b<p$, contains roughly $(b-a)/3$ cubic residues. The error terms have the order $\mathcal{O}(\sqrt p\cdot \ln p)$, so you still want $(b-a)$ to be large in comparison to $\sqrt{p}\cdot\ln p$ for this result to kick in. Anyway, it shows that the cubic residues are relatively evenly distributed. IOW not clustered in the sense @Conifold may have intended. $\endgroup$ – Jyrki Lahtonen Dec 2 '19 at 4:39
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One of your question is when we fix two (distinct cube free) positive integers $a,b$ and we choose randomly uniformly a prime $p\le N$ are the probabilities that $a$ and $b$ are cubes $\bmod p$ roughly equal (as $N$ gets large).

The answer is yes, because Chebotarev theorem tells us the asymptotic $$\#\{ p \le N, x^3-a\bmod p \text{ has a root }\}\sim \frac{|H|}{|G|} \frac{N}{\log N}$$ where $G=Gal(\Bbb{Q}(a^{1/3},\zeta_3)/\Bbb{Q}), |G|=6$ and $H$ is the set of $\sigma \in G$ such that $\sigma(a^{1/3})=a^{1/3}$, ie. $|H| = 2$ independently of $a$.

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