# Are cube roots evenly distributed modulo primes?

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.

• If $p\equiv1\pmod3$ then there are $\frac13(p-1)$ cubic residues modulo $p$. Does that answer your question? Dec 2, 2019 at 4:26
• $a^3\equiv b$ doesn't have solutions for all $b\pmod p$ Dec 2, 2019 at 4:28
• @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". Dec 2, 2019 at 4:29
• @J.W.Tanner: See my reply to LordSharktheUnknown... I'm really just trying to find when the roots exist modulo various $p$, $p$ prime. Dec 2, 2019 at 4:31
• 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. Dec 2, 2019 at 4:39

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 $$a$$.