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Solve $x^{15} \equiv 6 \pmod{7^2}$.

My approach based on Hensel's lemma:

First let's solve $x^{15} \equiv 6 \pmod{7}$, observe $3$ is a primitive root $\pmod{7}$, so let $x=3^y$ to get $15y \equiv 3 \pmod{6}$, solve to get $y_1=1, y_2=3, y_3=5$, the corresponding $x$'s are $x_1=3, x_2=6, x_3=5$

Now I use Hensel's lemma to lift the solutions to $\pmod{7^2}$. For example for $x_1$ one has $x_1^{15}-6=7 \times 2049843$ and $15x_1^{14}=3^{15} \times 5$, thus $7^0||(15x_1^{14})$, solve $2049843+3^{15} \times 5z_1 \equiv 0 \pmod{7}$ to get $z_1 \equiv 1 \pmod{7}$, so $w_1=x_1+7 \times z_1=10$ is a solution to the original equation.

Similar to $w_1$ one can lift $x_2,x_3$to get $w_2=6$ and $w_3=33$ are the other two solutions to the original equation $\pmod{49}$.

But does this way produce all the solutions?

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    For an example, where lifting fails (because the derivative will be divisiple by $p$ at a root) see [this question.](http://math.stackexchange.com/q/162344/11619)2012-08-17

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Yes, when Hensel's lemma applies, it gives all solutions.

One way to see why (the simplest form of) Hensel's lemma is true is nothing more than simply solving the equation

$ f(x + a p^k) \equiv 0 \pmod{p^{k+1}} $

for $a$ in a situation where $f(x) \equiv 0 \pmod{p^k}$ and the Taylor series for $f(x)$ makes sense and the higher order terms vanish, due to the high powers of $p$ involved:

$ f(x + a p^k) \cong f(x) + a p^k f'(x) \pmod{p^{k+1}} $

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    @user31899: Why wouldn't choosing $x_0\equiv w_0\pmod{p^k}$ work?2012-08-17