November 30th
Today I learned that for any polynomial $P(x_1,\ldots,x_n)\in\ZZ_p[x_1,\ldots,x_n]$ has a root in $(\ZZ_p)^n$ if and only if has roots in each $(\ZZ/p^\bullet\ZZ)^n.$ One direction of this easy. Indeed, if we have a root $(\alpha_1,\ldots,\alpha_n)\in(\ZZ_p)^n,$ then\[P(\alpha_1,\ldots,\alpha_n)=0\]in $\ZZ_p.$ Reducing all values$\pmod{p^\bullet}$ will give an equality there as well, so we get our solution in $(\ZZ/p^\bullet)^n$ by projecting $(\alpha_1,\ldots,\alpha_n).$
The harder direction is that if we have solutions for $P$ in $(\ZZ/p^\bullet\ZZ)^n$ for each $p^\bullet,$ then we can piece together a solution in $(\ZZ_p)^n.$ For this, we construct a Cauchy sequence in Hensel style. This is done in inductive-style steps. We construct a sequence of solutions\[(\alpha_{1,\bullet},\ldots,\alpha_{n,\bullet}),\]which holds in $(\ZZ/p^\bullet\ZZ)^n$ and satisfies $\alpha_{k,\bullet}\equiv\alpha_{k,\bullet+1}\pmod{p^\bullet}$; this of course makes a Cauchy sequence for each $\alpha_{k,\bullet}$ which converges to a solution $\alpha_k$ in $\ZZ_p.$ So this will be enough to give our solution in $(\ZZ_p)^n.$
The trick is that we're going to construct our solution such that, when reducing all of the solutions $(\ZZ/p^\bullet\ZZ)^n$ to our particular $(\ZZ/p^\bullet\ZZ)^n,$ infinitely many fall on top of $(\alpha_{1,\bullet},\ldots,\alpha_{n,\bullet}).$
As our base case, we begin with\[(\alpha_{1,0},\ldots,\alpha_{n,0})=(0,\ldots,0),\]which is a solution to $P$ in $(\ZZ/p^0\ZZ)^n.$ Further note that reducing any of our solutions to $P$ of $(\ZZ/p^\bullet\ZZ)^n$ into $(\ZZ/p^0\ZZ)^n$ will project here, so infinitely many lie on top of $(0,\ldots,0).$
Now suppose we have a solution $(\alpha_{1,\ell},\ldots,\alpha_{n,\ell})$ satisfying the statement. Reducing all of our solutions of $P$ in each $(\ZZ/p^\bullet\ZZ)^n$ into $(\ZZ/p^\ell\ZZ)^n,$ we have infinitely many that lie on top of $(\alpha_{1,\ell},\ldots,\alpha_{n,\ell}),$ by inductive hypothesis. Raising this into $(\ZZ/p^{\ell+1}\ZZ),$ note that there are only $p^n$ possible residues in $(\ZZ/p^{\ell+1}\ZZ)^n$ equivalent to $(\alpha_{1,\ell},\ldots,\alpha_{n,\ell}).$ However, infinitely many fall into this group of residue classes, so one of the $p^n$ residue classes must appear infinitely often. Name this class of $(\ZZ/p^{\ell+1}\ZZ)^n$ as\[(\alpha_{1,\ell+1},\ldots,\alpha_{n,\ell+1}).\]By construction, we have $\alpha_{k,\ell}\equiv a_{k+1,\ell}\pmod{p^\ell}$ for each $k,$ and infinitely many solutions fell on top of this residue class, making this a solution in $(\ZZ/p^\ell\ZZ)^n$ as well. This completes the inductive step.
This does complete the proof, with our Cauchy sequence in hand. What I find interesting about this proof is that it doesn't really use the fact that we're dealing with a polynomial anywhere in the proof. I guess implicitly used is that if\[P(\alpha_1,\ldots,\alpha_n)\equiv0\pmod{p^{\bullet+1}},\]then the equivalence holds in$\pmod{p^\bullet}$ as well, which is somewhat contingent on this being a polynomial. However, this is really it. Our polynomial can have as many variables as needed, it can be of any degree, it's coefficients can be anything from $\ZZ_p,$ and so on. It might even be possible to let these be power series, but this runs the risk of convergence problems.
Professor Kedlaya made the remark that this lemma is a bit misleading, in that it suggests that we need infinite information (solutions for each $\ZZ/p^\bullet\ZZ$) to get a solution in $\ZZ_p.$ Of course, this is not the case, a la Hensel's Lemma and the various ways to strengthen it. The prototypical example is quadratic residues, where, for $p \gt 2,$\[x^2\equiv c\pmod{p^\bullet}\]with $p\nmid c$ will have solutions if and only if $\left(\frac cp\right)=1.$ This is a finite check to get the infinite information that we need. The proof of this statement is exactly Hensel's Lemma, for the derivative of $x^2-c$ is $2x,$ the only roots of which are $0$ for $p=2.$ We have to push a little bit harder to get $p=2,$ I suppose, but it still only amounts to a finite check.