Let $K\mathbb{Q}$ be a number field, $D$ its discriminant and $\mathcal{O}$ the ring of integers in $K$. Let $x\in K$ (or maybe $\in \mathcal{O}[\frac 1D]$) such that for all primes $p$ in $\mathbb{Q}$ that split in $K$ and all $n\in\mathbb{N}$ we have the congruence $$ x^{p^{n1}}\equiv x^{p^n} \mod p^{2n} \mathcal{O}_p. $$ My question(s): which consequences does this have for $x$? Is $x$ a root of unity in $K$?
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$\begingroup$ What is $\mathcal{O}_p$? Is it $\mathcal{O}_K$ localized at $p$? $\endgroup$ – David Loeffler May 16 at 18:57

5$\begingroup$ Assuming that $vp$ is unramified, $x^{p^{n1}} \equiv x^{p^n} \mod v^{2n}$ and $(x,v) = 1$ implies that $p \log_v(x) \equiv \log_v(x) \mod v^n$. Taking $n$ arbitrarily large implies that $\log_v(x) = 0$ and $x$ is a root of unity. $\endgroup$ – Pound Sterling May 16 at 20:16

$\begingroup$ Yes, $\mathcal{O}_p$ is $\mathcal{O}_K$ localized at p. $\endgroup$ – LFM May 17 at 8:29