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I understand Hensel's Lemma and how you lift a root $f(b_{j}) \equiv 0 \pmod{p^\alpha}$ to one $f(b_{j+1}) \equiv 0 \pmod{p^{\alpha+1}}$, as long as $f'(b_{j}) \not\equiv 0 \pmod{p}$.

The equation is actually quite simple: $b_{j+1} \equiv b_{j} - f(b)*(f'(a)^{-1}) \pmod{p^{\alpha+1}}$.

My question is what happens when $f'(b) \equiv 0 \pmod{p}$.

The best I can find is that there are multiple solutions, but nothing I can find says how to find them. Is there a nice equation for $b_{j+1}$ (or the multiple values it can take on) given $b_{j}$.

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    I don't have my number theory books with me right now and I don't recall the discussion about this, but I remember that there are only two cases : either all integers lying above $b_j$ are roots, or either none of them are. There is no possibility that only 2 or 3 or ... out of $p$ are, it's all or nothing. If no one has answered this before I come home I'll quote one of my references down here with an answer.2011-12-12
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    Here is a more general form of Hensel's lemma: if $|f(a_0)|_p < |f'(a_0)|_p^2$ then there is a unique $p$-adic integer $a$ such that $f(a) = 0$ and $|a - a_0|_p < |f'(a_0)|_p$. This is more important than trying to find a multi-valued formula for all solutions of $f(x) \equiv 0 \bmod p^n$.2011-12-12
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    Niven/Zuckerman/Montgomery should have this information, IIRC.2011-12-12
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    @Dylon Moreland: The WP link you posted actually starts *editing* the section in question. If many people click on it, somebody might actually destroy the article section accidentally!2011-12-12
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    @Marc Woops! [Here's the correct link](http://en.wikipedia.org/wiki/Hensel%27s_lemma#Hensel_Lifting).2011-12-12

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