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Let $f(n)$ denote a cubic polynomial in $n$ with integer coefficients. Fix a prime $p$. Suppose I know $\#\{a \pmod{p} : f(a) \equiv 0 \pmod{p}\}$. Can one determine $\#\{a \pmod{p^2} : f(a) \equiv 0 \pmod{p^2}\}$?

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If $f(x)=pg(x)$, then $f=0 \mod p$, but what can we say $\{a:g(a)=0 \pmod p\}$? It seems there are not many informations..Just a comment. – wxu Mar 17 '13 at 2:43
up vote 4 down vote accepted

For any prime $p$ that doesn't divide the discriminant of $f$, the two quantities are the same. In other words, of the $p$ possible "lifts" of $a\pmod p$ - namely $a, a+p, a+2p, \dots, a+(p-1)p\pmod{p^2}$ - exactly one of them will be a root of $f$ modulo $p^2$. This follows from Hensel's lemma, which can also treat the finitely many primes that do divide the discriminant of $f$.

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