$p$-adic expansion

Question

I have just touched on this topic, please guide me along. If I have a prime number $p=10^{10}+19$, and a $p$-adic number $\alpha=\frac{16}{17}$. How do I derive its $p$-adic expansion?

Thanks in advance.

Answer 1

Suppose $-1< \frac{a}{b}\le0$ is rational with $p\nmid b$. By Fermat's little theorem, $b\mid(p^n-1)$ for some $n$.

Write $\displaystyle\frac{a}{b}=\frac{a_0+a_1p+\cdots+a_{n-1}p^{n-1}}{1-p^n}$ with each $0\le a_i<p$. This expands out to be

$$a_0+a_1p+\cdots+a_{n-1}p^{n-1}+a_0p^n+a_1p^{n+1}+\cdots+a_{n-1}p^{2n-1}+a_0p^{2n}+\cdots$$

by the geometric sum formula. Any $x\in\Bbb Q$ can be written as $n+r$ with $n\in\Bbb Z$ and $-1<r\le0$, and if $p\mid a$ or $p\mid b$ we can shift the digits (coefficients of powers of $p$) right or left appropriately.

Write $\frac{16}{17}=1-\frac{1}{17}$. We compute $b=10^{10}+19\equiv(-2)^5+2\equiv4$ mod $17$. As $4^2\equiv-1$ we find that this has order $4$. Now, $p=10^{10}+19=10000000019$ With Alpha we can compute

$$\cdot\frac{(10^{10}+19)^4-1}{17}=588235298588235306858823545550588242960$$

and then 588235298588235306858823545550588242960 in base 10000000019 returns the four digits 588235295:2352941180:9411764723:7647058838. Reverse this sequence (since the $p$-adic expansion goes from left to right instead of right to left), make it repeating over and over again, and then finally add $1$ to the first digit:

$$\begin{array}{lll} & 7647058839 &+ & 9411764723p & + & 2352941180p^2 & + & 588235295p^3 \\ + & 7647058838p^4 &+ & 9411764723p^5 & + & 2352941180p^6 & + & 588235295p^7 \\ + & 7647058838p^8 &+ & 9411764723p^9 & + & 2352941180p^{10} & + & 588235295p^{11} & +~~\cdots\end{array}$$