Show that $y^2=x^3+1$ has infinitely many solutions over $\mathbb Z_p$. I first compared it with how I would solve this over the real numbers. You would  say:


*

*$y^2=\alpha$ has a solution for all $\alpha>0$, of which there are infinitely many. 

*$x^3+1>0$ for all $x>-1$, of which there are also infinitely many.


However I can't seem to extend this way of thinking to $\mathbb Z_p$. I have a strong hunch that I need to use Hensel's Lemma in some way, but I just can't see how. 

Instead of appealing either to the Binomial Theorem, as Qiaochu did, or to Hensel, you can do it “analytically”, knowing that in terms of the local uniformizing parameter $x$ at $(0,1)$, you must be able to expand $\eta=y-1$ as a series in $x$ with no constant term. Making that substitution, you get from $y^2=x^3+1$ to $\eta=(x^3-\eta^2)/2$, which you can look at as a recursive procedure for getting the expansion of $\eta$. The result is
$$
\eta=\frac12x^3 - \frac18x^6 + \frac1{16}x^9 - \frac5{128}x^{12} + \frac7{256}x^{15}-\cdots\,,
$$
and of course this is exactly the Binomial expansion. Just make a substitution $x\mapsto x_0$ for $x_0$ small enough, and you’ll get a convergent series.
I should say that this procedure is much more fun if you do it in the neighborhood of the point $\Bbb O$ at infinity of this curve, that is, $(0:1:0)$ projectively. There, the description of the curve in terms of $\xi=x/y$  and $\zeta=1/y$ is $\zeta=\xi^3+\zeta^3$, the l.u.p. being $\xi$, so you get a $\Bbb Z$-series for $\zeta$ in terms of $\xi$. Far as I can see, this expansion doesn’t come out of the Binomial Theorem.
