Finding number of roots of a polynomial in p-adic integers $\mathbb{Z}_{p}$ The problem is to find the number roots of $x^3+25x^2+x-9 $ in $\mathbb{Z}_{p}$ for p=2,3,5,7.
I read this equivalent to having a root mod $p^{k}~\forall k\geq 1$.
By Newton's lemma I can get whether there is at least one root.
Any suggestions on how to find the number of roots?
 A: $5$ and $7$ should be straightforward, $3$ is special because of the degree $3$ and $2$ is always special, isn't it?


*

*You will notice by inspection that there are no solutions $\pmod 4$, even though $1$ is a solution $\pmod 2$.

*Above $0\pmod 3$, there is only $0\pmod 9$, above that only $9\pmod {27}$, and so on using Hensel; above $1\pmod 3$, there are $1,4,7\pmod 9$. Recapitulate: If $x^3+25x^2+x-9\equiv 0\pmod{3^k}$ with $k\ge1$, then $$(x+3^ki)^3+25(x+3^ki)^2+x+3^ki-9\equiv (50x+1)3^ki+(x^3+25x^2+x-9)\pmod{3^{k+1}},$$ so there is exactly one sutable $i\in\{0,1,2\}$ provided $50x+1\not\equiv 0\pmod 3$. That's why $0\pmod 3$ gave us exactly one solution in $\mathbb Z_3$. And that's also why $1\pmod 3$ could create more solutions $\pmod 9$. When you go to higher powers of $3$, you will notice that history repeats: you will always have three solutions $\pmod {3^k}$, where two lead to unique solutions $\pmod {3^{k+1}}$ and then unique solutions in $\mathbb Z_3$, whereas the third produces three fresh solutions $\pmod{3^{k+1}}$ again; ultimately this gives you infinitely many solutions in $\mathbb Z_3$!

*By insepection, there is no solution $\pmod 5$

*There is a unique solution $6\pmod 7$, which gives us a unique solution in $\mathbb Z_7$ by Hensel.
A: 1)Using Newton's polygon method
2)Hensel's lemma
