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Suppose that we are given a fixed pair $a_1, a_2$ of non-zero irrational algebraic integers in some number field $K$ which are independent over $\mathbb{Q}$. Suppose that $\mathcal{P}$ is a prime ideal in $\mathcal{O}_K$, the ring of integers in $K$. For which $\mathcal{P}$ can we find a rational number $b_1/b_2$ say such that $a_1/a_2 \equiv b_1/b_2 \pmod{\mathcal{P}}$?

A second question: Suppose we have two linear forms with algebraic integral coefficients in some number field $K$, linearly independent over $\mathbb{Q}$,

$$\displaystyle a_1 x_1 + a_2 x_2 + a_3 x_3, \text{ } b_1 x_1 + b_2 x_2 + b_3 x_3. $$

Is there a way to estimate the number of solutions of the following congruence

$$\displaystyle a_1 x_1 + a_2 x_2 + a_3 x_3 \equiv b_1 x_1 + b_2 x_2 + b_3 x_3 \equiv 0 \pmod{\mathcal{P}}$$

for prime ideals $\mathcal{P}$ which can be assumed to be sufficiently large in terms of $a_i$'s and $b_i$'s? For a given $\mathcal{P}$, define $E_\mathcal{P}(X)$ to be the number of solutions of the above congruence in integer triplets $(x_1, x_2, x_3) \in \mathbb{Z}^3$ with $|x_i| \leq X$ for $ i = 1,2,3$. Further, we may assume that the rational prime $p$ below $\mathcal{P}$ satisfies $p > X$. Can we obtain a bound of the form

$$\displaystyle \sum_{\substack{\mathcal{P} \\ p > X}} E_\mathcal{P}(X) = o(X^3)?$$

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    $\begingroup$ The first question is straightforward. Let $p$ be the rational prime below $\mathcal{P}$. You can find a rational number $q$ so that $a_{1}/a_{2} \equiv q \pmod{\mathcal{P}}$ if and only if $a_{1}/a_{2} \in \mathbb{F}_{p} \subseteq \mathcal{O}_{K}/\mathcal{P}$. For the second question, where do the variables $x_{i}$ live? Are they in $\mathcal{O}_{K}$? If so, what do you mean by $|x_{i}| \leq X$? $\endgroup$ – Jeremy Rouse Nov 18 '14 at 18:09
  • $\begingroup$ My apologies, I thought I had stated that $x_i$ \in $\mathbb{Z}$ $\endgroup$ – Stanley Yao Xiao Nov 18 '14 at 20:17

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