Solve $x^3 - x - 1=0$ over $\mathbb{R} \times \mathbb{Q}_7$ Is possible to find a solution to $x^3 - x - 1 \approx 0$ using p-adic numbers? I can state this question as two inequalities.  Find $x = \frac{m}{n} \in \mathbb{Q}$:

*

*$|x^3 - x - 1 |_\infty < 0.001$


*$|x^3 - x - 1 |_7 < \frac{1}{7^3}$
There are infinitely many such fractions.  So another more general question could be to describe them.  How about, let's limit ourselves to $x = \frac{m}{n}$ with $|m|+|n| < 10^6$.  I would accept a computer solution if there are too many solutions.
We could write the simultaneous inequalities as, solve for $x = \frac{m}{n} \in \mathbb{Q}$ with $|m| + |n| < 10^6$ ("taxicab norm" or $L^1$ norm or $\ell_1$ distance even though there's only two coordinates.)

*

*$|x^3 - x - 1 | < \frac{1}{10^3}$


*$x^3 - x - 1  \equiv 0 \pmod {7^3}$
 A: Here's a fairly generic method, start by using Newton's method to get a solution to arbitrary accuracy in both fields. Let's suppose our approximations are the rational numbers, $x_\infty$ and $x_7$.
Now we can join them into a single solution by a linear combination of both of them, $$x=ax_7 +b x_\infty$$
It's a fun exercise to see that the following two sequences of rational numbers have limits that go to 0 in one field and 1 in the other, for our purposes an easy choice is for $n$ large as you like,
$$a = \frac{1}{1+7^n}$$
$$b = \frac{7^n}{1+7^n}$$
So this means
$$x=\frac{x_7+x_\infty 7^n}{1+7^n}$$
So here as an example, just recklessly picking a bunch of digits arbitrarily $x_\infty = \frac{1324717957244746}{10^{15}}$ and $x_7 = 67770352087443486$ and $n=40$ we have,
$$x = \frac{4217110960882744163328506069361294227335237174373}{3183402880454513992870717569612001000000000000000}$$
This happens to be close up to about $17$ digits in the reals and $20$ digits in the $7$-adics. I didn't count the digits of accuracy for my example beforehand, but this will just be limited by whatever point you stopped your digits of accuracy in the local fields since the $a,b$ sequences can always be chosen arbitrarily close to $0$ and $1$.
