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From the topic , how many prime $p$ that cannot find any integer $x$ to satisfy $$x^3-2028x+2018\equiv 0 \pmod{p^3}$$

I try to start with $x^3-2028x+2018\equiv 0 \ \ \left (mod \ p \right )$ and put some small $p$ like $2,3,5,7$ and then notice which $x$ can be applied , it's tedious task for me to do with more bigger $p$.

But I don't know how to deal with $p^3$ and I'm not sure that there are infinite $p$ to match at least one $x$ or not?

i'm appreciate that you could hint me for some theory to start with. Thank you to any reply.

gt6989b
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ABCDEFG user157844
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  • Where did you get this question from? What is the most recent thing you studied prior to encountering this? Hensel lemma, for example? – Sarvesh Ravichandran Iyer Oct 24 '19 at 04:58
  • I create any random question with my friend and my friend argues to me. He said that there are infinite $p$ that cannot find any $x$ to satisfy this, and yes, I've read about Hensel's lemma but don't understand enough. That leads to this question. – ABCDEFG user157844 Oct 24 '19 at 05:03
  • Ouch. Full marks for creating the question, but this one may be a step too far. Read up on cubic congruences. – Sarvesh Ravichandran Iyer Oct 24 '19 at 05:06
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    The appearance of 2018 suggests a contest from last year. Anyway, Hensel's lemma will usually allow you to lift a solution mod $p$ to a solution mod $p^3$. The possible exceptions being $2,3,5,7,59$ and $16567$. Probably some of those can still be handled by being careful about the choice of root to be lifted. Solvability modulo $p$ OTOH... – Jyrki Lahtonen Oct 24 '19 at 05:10
  • @JyrkiLahtonen It's not a contest question, I've created this question by myself indeed, I swear. – ABCDEFG user157844 Oct 24 '19 at 05:13
  • No problem. Even if it were a contest question, it would obviously be past the deadline :-) – Jyrki Lahtonen Oct 24 '19 at 05:14
  • @JyrkiLahtonen Thank you, I'm not knowing much about those Lemma, I look forward to the solution from someone who clarifies this. – ABCDEFG user157844 Oct 24 '19 at 05:16
  • Anyway, the cubic is irreducible over $\Bbb{Q}$, and its discriminant (factors listed above) is not a square, so the Galois group is $S_3$. As that is non-abelian, we cannot expect an answer of the form saying that it works for primes $p$ in certain residue class modulo some number $m$. It may be possible to list finitely many binary quadratic forms, and say that solutions modulo $p^3$ exist iff $p$ can be written as a value of one of those forms. Unfortunately I'm not familiar with that theory. If Will Jagy shows up, and can spare the cycles... Or may be I'm ignorant :-/ – Jyrki Lahtonen Oct 24 '19 at 05:26
  • @JyrkiLahtonen I wonder if we can find some $x$ to satisfy $x^3 - 2028x+2018\equiv 0 \ (mod \ 16567) $ by hands, or using any trick ? (e.g.derivative) – ABCDEFG user157844 Oct 24 '19 at 05:35
  • Because $16567$ is a factor of the discriminant, the cubic has a double root. Implying that $x^3-2028x+2018$ has a common factor with its derivative $3x^2-2028$ modulo $16567$. You can calculate the gcd of two low degree polynomials relatively easily by hand with Euclid's algorithm. – Jyrki Lahtonen Oct 24 '19 at 07:04
  • For a little bit of material about algorithms for solving polynomial congruences modulo a large prime see this thread. Alas, I could only squeeze in a toy example with a small prime, and such algorithms don't really help here. – Jyrki Lahtonen Oct 24 '19 at 07:09

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Very generally, if $f$ is a nonlinear irreducible polynomial over $\mathbb{Z}$, then there are always infinitely many prime numbers $p$ such that $f$ has no roots mod $p$. By the Frobenius density theorem (which you can find an excellent brief overview of here), it suffices to show that some element of the Galois group $G$ of $f$ has no fixed points (when acting on the roots of $f$). But the set of elements of $G$ that have a fixed points is just the union of the conjugates of the subgroup $H$ that fixes one particular root, since $G$ acts transitively on the roots. Since a finite group cannot be covered by the conjugates of a proper subgroup, some element of $G$ has no fixed points.

In particular, then, your cubic $f(x)=x^3-2028x+2018$ is irreducible over $\mathbb{Z}$ since it is monic and has no integer roots, so there are infinitely primes $p$ such that it has no roots mod $p$, and hence also no roots mod $p^3$.

There might be a more elementary way to prove this for this particular polynomial, but if it's just some random polynomial you chose, I wouldn't expect there to be one.

Eric Wofsey
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