Solve $x^3 - x - 1=0$ over $\mathbb{R} \times \mathbb{Q}_7$

Question

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.)

• $|x^3 - x - 1 | < \frac{1}{10^3}$

• $x^3 - x - 1 \equiv 0 \pmod {7^3}$

Answer 1

I arrive at $7193/5430=1.32467...\equiv1384\bmod 2401$. The base ten representation is close enough to the exact answer (1.32471...) to guarantee $|x^3-x-1|<0.001$, and $1384\bmod2401$ gives $x^3-x-1\equiv0$ exactly.

I will first seek a solution for $|x^3-x-1|<0.001$, using the real-arithmetic norm, via continued fractions. This in itself does not solve the simultaneous equations, but it can be used as a springboard to such a simultaneous solution.

Start with the fact that by substitution we find there is a root $r_0$ between $1$ and $2$ (which is the only real root, actually, as we can infer from the cubic discriminant $27(-1)^2+4(-1)^3>0$). So render this root as the continued fraction $[1,r_1]=1+(1/r_1)$. Then

$(1+(1/r_1))^3-(1+(1/r_1))-1=0$

Expanding the binomial power $(1+(1/r_1))^3$, then clearing fractions and collecting terms gives

$r_1^3-2r_1^2-3r_1-1=0$

with its root between $3$ and $4$. Thus $r_0=[1,3,r_2]$ with $r_1=3+(1/r_2)$. Thereby

$(1+(1/r_2))^3-2(1+(1/r_2))^2-3(1+(1/r_2))-1=0$

$r_2^3-12r_2^2-7r_2-1=0$

We find $r_2$ lies between $12$ and $13$.

Therefore our root lies between $[1,3,12]=49/37=1.32432...$ and $[1,3,13]=53/40=1.325$. These bounds lie within $0.001$ of each other, but it turns out that neither bound actually satisfies the condition $|x^3-x-1|<0.001$. Although the true root itself is within $0.001$ of the bounds, the derivative at the root is greater than $1$ and thus the error in the polynomial value is greater.

There is, however, a subtle way to improve those bounds. If we go back to the cubic equation for $r_2$ we note that since $r_2$ is clearly greater than $1$ we have

$r_2^3-12r_2^2-7r_2-r_2<r_2^3-12r_2^2-7r_2-1<r_2^3-12r_2^2-7r_2-0$

$r_2(r_2^2-12r_2-8)<r_2^3-12r_2^2-7r_2-1<r_2(r_2^2-12r_2-7)$

Thus the true root $r_2$ lies between the roots of the quadratic equations $r_2^2-12r_2-7=0$ and $r_2^2-12r_2-8=0$, thus between $6+\sqrt{43}$ and $6+\sqrt{44}$. As $(13/2)^2=42.25$ and $(20/3)^2=44.44444...$ this implies that $r_2$ actually lies between $25/2=[12,2]$ and $38/3=[12,1,2]$. Thus

$[1,3,12,2]<r_0<[1,3,12,1,2]$

$102/77<r_0<155/117$

These bounds do give $|x^3-x-1|<0.001$, and in particular the fraction $102/77$ is the minimal such fraction in terms of numerator and denominator size.

To bring in the $7-adic$ requirement we must first find the solution in $7$-adics alone. Modulo $7$ the equation $x^3-x-1=0$ has the unique solution $x\equiv5$, and by Hensel lifting this can be converted to $x\equiv1384\bmod2401$. The latter, represented in $7$-adics as $...4015$, implements the requirement $|x^3-x-1|_7\le 7^{-4}<7^{-3}$.

We then take the bounds derived from the continued fraction above and form a combined fraction lying between them:

$\dfrac{102a+155b}{77a+117b}$

where $a$ and $b$ are positive whole numbers and the $7$-adics requirement renders

$102a+155b\equiv 1384(77a+117b)\bmod 2401$

$-106466a\equiv161773b\bmod 2401$

$1579a\equiv906b\bmod2401$

$a\equiv 69b\bmod 2401$

So simply render $a=69,b=1$:

$x=\dfrac{102×69+155}{77×69+117}=\color{blue}{\dfrac{7193}{5430}}.$

Answer 2

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 $1$.