For what $n$ does $\left\lvert\frac{p^nx-1}{p+1}\right\rvert_p=1$, given $\lvert x\rvert_p=1$?

Question

For what $n$ does $\left\lvert\dfrac{p^n\cdot x-1}{p+1}\right\rvert_p=1$, given $\lvert x\rvert_p=1$?

Where $\lvert x\rvert_p$ is inverse of the highest power of $p$ that divides $x$ (i.e. the p-adic absolute value)

---

My (wrong) attempt, I have now deleted.

Answer 1

For all $n>0$. Recall that $|a\pm b|_p=\max\{|a|_p,|b|_p\}$ if $|a|_p\ne|b|_p$. Then as $|p^nx|_p=p^{-n}$, we have $$\left|\frac{p^nx-1}{p+1}\right|_p =\frac{|p^nx-1|_p}{|p+1|_p}=\frac{\max\{p^{-n},1\}}{\max\{p^{-1},1\}}=1.$$ For $n<0$, we arrive at $$\left|\frac{p^nx-1}{p+1}\right|_p =\frac{\max\{p^{-n},1\}}{\max\{p^{-1},1\}}=p^{-n}$$ instead, so this is not a solution.

For $n=0$, the answer depends on $|x-1|_p$, i.e., whether $x\equiv 1\pmod p$.