Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\#\{a \pmod{p^2} : f(a) \equiv 0 \pmod{p^2}\}$?

share|improve this question
1  
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

1 Answer 1

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 $f$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.