Solve $x^{2} = -1$ in $\mathbb{Q}_{5}$.

Question

I'm trying to find the $p$-adic representation of the solution of $x^{2} = -1$ in $\mathbb{Q}_{5}$. I'm just starting to learn about the $p$-adic numbers and fields so I don't even really know where to begin (I've been having a bit of a hard time grasping what $p$-adic numbers really even are). Any hints/help would be appreciated, thanks!

Edit: I think I may have found a way to approach the problem; I've tried the following:

Let $f(x) = x^{2} + 1 \equiv 0 \,(\text{mod} \, 5)$. This has two solutions, namely $x \equiv 2 \, (\text{mod} \, 5)$ and $x \equiv 3 \, (\text{mod} \, 5)$. We start with $x \equiv 2 \, (\text{mod} \, 5)$. Then,

\begin{align*} f(2+5t) &\equiv 20t+5\,(\text{mod} \, 5^{2}) \implies t \equiv 1\, (\text{mod} \, 5)\\ f(2+1(5)+5^{2}t) &\equiv 350t+50 \, (\text{mod} \, 5^{3}) \implies t \equiv 2\,(\text{mod} \,5)\\ f(2+1(5)+2(5^{2}) + 5^{3}t) &\equiv 14250t+3250 \, (\text{mod} \, 5^{4}) \implies t \equiv 1 \, (\text{mod} \, 5)\\ \end{align*} and so on, with each successive step yielding the next digit of the $p$-adic representation. So, the first solution would be represented by $x = 2 + 1(5) + 2(5^{2}) + 1(5^{3})+...$

Is this a reasonable way to approach this?

Answer 1

Just to show how to adapt two methods from How to compute 2-adic square roots?:

Newton-Raphson method:

Works exactly as known from calculus, if the start value is good enough. Iteration formula is $x_{n+1} = \dfrac{x_n}{2} -\dfrac{1}{2x_n}$. Let's take $x_0=2$, then

$x_1=\frac34, x_2=-\frac{7}{24}, x_3= \frac{527}{336},...$

Does this look good? Well not on first sight, our eyes have to get used to $5$-adic thinking. But look:

$\lvert(\frac34) ^2 -(-1)\rvert_5 = 1/25 =5^{-2}$

$\lvert(\frac{-7}{24}) ^2 -(-1)\rvert_5 = 1/625 = 5^{-4}$

$\lvert(\frac{527}{336}) ^2 -(-1)\rvert_5 = 1/390625 = 5^{-8}$ ...

... yes, those fractions converge to a square root of $-1$. The hardest part here would be to find their $5$-adic representations. Well,

$\frac34 = 1+ \frac{1}{1-5} = 1+ (1 + 5 + 5^2 + 5^3 + ...) = (2 + 1\cdot 5 + 1\cdot 5^2 + 1 \cdot5^3 + ...) = ...111112_5$

$\frac{-7}{24} = 12_5 \cdot \frac{1}{1-5^2} = 12_5\cdot (..10101_5) = ...121212_5$

$\frac{527}{336}= $ well, that needs some work, but it's $...431212_5$.

---

Binomial series:

Namely, $(1+x)^\frac{1}{2} = 1 +\frac12 x -\frac18 x^2 + \frac{1}{16}x^3 -\frac{5}{128} x^4 + ... +\binom{1/2}{k}x^k + ...$

which converges $5$-adically as soon as $\lvert x\rvert_5 \le \frac15$. The problem is that if we are too impatient and want to try that the square root of $-1$ directly, it won't work, because then $x=-2$ which is not small enough $5$-adically. Luckily, we can pull out a factor of $4$ though, and get

$\sqrt{-1} = \sqrt{4 \cdot (-\frac14)} = 2 \cdot \sqrt{-\frac14} = 2 \cdot \left(1 + (-\dfrac54 )\right)^\frac{1}{2}$

and since $\lvert -\dfrac54 \rvert_5 = \frac15$, the above series will converge. Up to the fourth power term, we have

$\left(1 + (-\dfrac54 )\right)^\frac{1}{2} \approx_{5-\text{adically}} \dfrac{-1237}{32768}$

i.e. $\sqrt{-1} \approx_{5-\text{adically}} \dfrac{-1237}{16384}$

which is $...31212_5$, and here these $5$-adic expansions are a bit easier to calculate because the denominator of the fraction contains only powers of $2$, so we only need to calculate those modulo $5^k$, which is relatively easy.

Added: To find the $p$-adic expansions of rational numbers, cf. Finding p-adic expansions of rational numbers.