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Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial and let $p$ be a prime number. I'm looking for results about $$N_f(p^k) := \#\{(f(n) \bmod p^k) : n \in \mathbb{Z}\},$$ as $k \to +\infty$, where $(a \bmod b)$ is the remainder of $a$ divided by $b$.

If $N_f(p) = p$ and $f^\prime(x) \not\equiv 0\bmod p$, for all $x \in \mathbb{Z}$, then using Hensel's lemma it can be proved that $N_f(p^k) = p^k$ for all $k \geq 1$. However, I did not find more general result.

Thank you for any suggestion.

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  • $\begingroup$ people.maths.bris.ac.uk/~mazag/nt/lecture7.pdf. Last page tells you an explicit of lifting in general. Also Chebotarev theorem tells you the density of primes for which $N_f(p)$ is non-empty. $\endgroup$
    – Vlad Matei
    Mar 16, 2018 at 17:49
  • $\begingroup$ Chebotarev is only useful here if $p$ is large compared to the degree of $f$. In this case, there is even an asymptotic formula for the size of $N_f(p)$ in terms of the two Galois groups of $f(x)-t$ over $\mathbb F_p(t)$ and $\bar{\mathbb F}_p(t)$, respectively. $\endgroup$
    – Peter Mueller
    Mar 16, 2018 at 18:01
  • $\begingroup$ @VladMatei That lecture maybe useful. Anyway, $N_f(p)$ is always non-empty (actually non-zero) I don't know what you meant. $\endgroup$
    – OhhiMark
    Mar 16, 2018 at 20:27
  • $\begingroup$ The result Peter Mueller refers to (asymptotic as p^k tends to infinity) turns out to be a theorem by Birch and Swinnerton-Dyer doi.org/10.4064/aa-5-4-417-423 . I can't say it was very obvious to locate. $\endgroup$
    – Sary
    Oct 27, 2022 at 13:30

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At this question, it was proven that for all $f$, there exists a rational number $\delta$ such that $N_f(p^k)\sim \delta\cdot p^k$ as $k\rightarrow\infty$.

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