Say K = $\mathbb{Q} (\sqrt{m})$ be a quadratic number field, R = $\mathcal{O}_{L}$ be its ring of integers. We say a prime 'p' is odd if and only if 'p' does not show up in factorization of the ideal $\langle 2 \rangle$ in R.
In integers, we know that for an odd prime 'p', a number 'a' is a square modulo 'p' if and only if a is a square modulo $p^{k}$ for any $ k \geq 1 $.
Does analog of this hold in R i.e. given any odd prime 'p', any $ a \in R $ is a square modulo the ideal $\langle p \rangle$ if and only if a is square modulo $\langle p \rangle^{k}$ for any $ k \geq 1 $ ?