# Is an algebraic number satisfying certain super-congruences a root of unity?

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 $$x^{p^{n-1}}\equiv x^{p^n} \mod p^{2n} \mathcal{O}_p.$$ My question(s): which consequences does this have for $$x$$? Is $$x$$ a root of unity in $$K$$?

• What is $\mathcal{O}_p$? Is it $\mathcal{O}_K$ localized at $p$? – David Loeffler May 16 at 18:57
• Assuming that $v|p$ is unramified, $x^{p^{n-1}} \equiv x^{p^n} \mod v^{2n}$ 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. – Pound Sterling May 16 at 20:16
• Yes, $\mathcal{O}_p$ is $\mathcal{O}_K$ localized at p. – LFM May 17 at 8:29