This is about p-adic norm. When $\left|\left(\frac{u}{v} \right)^N\pm 1 \right|_p<1$?

Question

This is about p-adic norm.

If $x=[\left(\frac{u}{v} \right)^N\pm 1], \ u,v \in \mathbb{Z}, \ N \in \mathbb{N}$, then which conditions on $u,v, N$ ensures that $\left|\left(\frac{u}{v} \right)^N\pm 1 \right|_p<1$ ?

Actually my purpose is to show when $(1 +x)^r$ or $(1-x)^r, \ r \in \mathbb{Q}$ is a rational number given that $|x|_p<1 $.

If I take $ x=[\left(\frac{u}{v} \right)^N\pm 1]$, then clearly $(1\pm x)^{r=\frac{1}{N}} \in \mathbb{Q}$ but at the same we need to make sure whether $\left|\left(\frac{u}{v} \right)^N\pm 1 \right|_p<1$ .

Kindly help me

Answer 1

To answer the question posed here: We have $\vert (\frac{u}{v})^N \pm 1\vert_p < 1 \Leftrightarrow \vert u^N \pm v^N\vert_p < \vert v^N \vert_p$ which with standard arguments implies $\vert u \vert_p = \vert v \vert_p$. Assuming, as one should, $gcd(u,v) =1$, that common $p$-adic value is $= 1$, and the original pair of inequalities is equivalent to $u^N \equiv \mp v^N$ mod $p$.

This is just a question about the residues $\bar u, \bar v$ of $u,v$ in the finite field $\Bbb Z /p$ and can be reformulated as follows: Since w.l.o.g. we assume that $u,v$ are not divisible by $p$, we have $\bar u, \bar v \in (\Bbb Z/p)^\times$, so $\bar u\cdot \bar v^{-1} \in (\Bbb Z/p)^\times$, and the question is whether the $N$-th power of that element is $\mp 1$, or in yet other words:

the original "$+$" case is true iff the order of $\bar u\cdot \bar v^{-1} \in (\Bbb Z/p)^\times$ is exactly $2N$

(unless $p=2$, in which it just needs to be a divisior of $N$), whereas

the original "$-$" case is true iff the order of $\bar u\cdot \bar v^{-1} \in (\Bbb Z/p)^\times$ divides $N$.

So e.g. the original "$-$" case is true for all $u, v$ (not divisible by $p$) if $N$ happens to be a multiple of $p-1$. As another example, the original "$-$" case is true for any $N$, and the original "$+$" case at least for any odd $N$, if $u \equiv v$ resp. $u \equiv -v$ mod $p$.

To actually compute the order of elements in finite fields is not an easy problem in general, but in some small examples can be done by hand (and in reasonably small ones with a computer). Compare e.g. Order of elements in finite fields, Fastest Way to Find order of element in Finite Fields?, Easy way to find the order of elements in a finite field, Determining orders of elements in extensions of finite fields, Element of given order in a finite field.

---

Finally, let me note that I am not sure if your proposed "answer" in the duplicate of this question makes sense, and if you interpret the answers and comments in Shw that $ \ (1+X)^a \in \mathbb{Q}$ both in $\mathbb{R}$ and $ \mathbb{Q}_p$ under some condition correctly by asking this exact question.