I want to determine the number of solutions of $f(x)=x^{19}-3x+2=0$ over $\mathbb{Z}_{19}$ and $\mathbb{Z}_{17}$ ($p$-adic integers).

Is the following strategy correct for $\mathbb{Z}_{19}$ (and analogously for $\mathbb{Z}_{17})$? Check for all integers $n\in\{0,\dotsc,19-1=18\}$ whether $f(n)\equiv 0\mod 19$ and $f'(n)\not\equiv 0\mod 19$ both hold. Then Hensel's lemma will guarantee for each such $n$ that there exists a unique $b\in\mathbb{Z}_{19}$ such that $f(b)=0$ and $b\equiv a \mod p$. Thus, the number of solutions of $f(x)=0$ is exactly the number of possible values for $n$ that satisfy the two aforementioned equivalences.

  • $\begingroup$ But an integer $n$ with $f(n)\equiv0\pmod{19}$ may also lift to a solution in $\Bbb{Z}_{19}$ even if $f'(n)\not\equiv0\pmod{19}$. $\endgroup$ Jun 22, 2019 at 18:48
  • 2
    $\begingroup$ @Servaes: Do you mean "even if $f'(n) \equiv 0$ (mod $19$)"? $\endgroup$ Jun 22, 2019 at 22:35
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    $\begingroup$ @TorstenSchoeneberg Oh yes indeed I do! Too late to edit unfortunately. $\endgroup$ Jun 22, 2019 at 23:00

1 Answer 1


Your strategy is essentially right, and will work here (you are tacitly using if $f(n) \not \equiv 0$ mod $p$ for an $n \in \lbrace 0, 1, ..., p\rbrace$, then $f(x) \neq 0$ for any $x \equiv n$ mod $p$, and every $x\in \mathbb Z_p$ is $\equiv n$ for a unique such $n$).

One should add the following caveat:

As user Servaes points out in a comment, a potentially tricky case is $f(n) \equiv f'(n) \equiv 0$ mod $p$. Here, everything can happen, i.e. $f$ could have no, one, or more roots $\equiv n$ in $\mathbb Z_p$ (examples: $p=2$ and $f(x) = x^2-2$, resp. $x^2$, resp. $x^2-1$).

However, you are lucky since for your $f$ we have $f'(x) \equiv -3 \not \equiv 0$ mod $19$, and $$f'(x) \equiv 19x^{18}-3 \equiv 2x^2-3 \text{ mod } 17,$$ which is easily seen to have no zeroes mod $17$ either.


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