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The following problem came up on a mailing list that I subscribe to:

If $\alpha$ is irrational we can find (using continued fractions) infinitely many rational fractions $p/q$ such that $|q \alpha - p| \le C q^{-1}$ for some absolute constant $C >0$. However, suppose that one also restricts $p/q$ to be the norm of an algebraic number in $\mathbb{Q}(\sqrt{-1})$, is the statement still true? If so, is there a good algorithm to find all such?

Of course one can ask this where $\mathbb{Q}(\sqrt{-1})$ is replaced by any number field. I know that there a number of results about approximation by elements of $\Gamma$ a finitely generated subgroup of $\mathbb{C}$ but that doesn't apply here.

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  • $\begingroup$ What does the Oppenheim conjecture (proved by Margulis) give when applied to $x^2+y^2-\alpha(z^2+w^2)$? $\endgroup$ – Felipe Voloch Aug 14 '13 at 23:44
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    $\begingroup$ Should be true typically but not always. If $\alpha\bmod 1$ is chosen randomly, then with probability $1$ the best $q$ grow exponentially, and the chance that a random number around $C^n$ is the sum of two squares is about $c/\sqrt{n}$, whose sum diverges. But it seems you can use "Chinese Remainder" to recursively concoct a continued fraction with fast-growing denominators that converges to some $\alpha$ for which $\liminf_{q\rightarrow\infty} q \left\| \alpha q \right\| = \infty$. $\endgroup$ – Noam D. Elkies Aug 14 '13 at 23:48
  • $\begingroup$ @Noam: Thanks. The paper matwbn.icm.edu.pl/ksiazki/aa/aa53/aa5323.pdf by Glyn Harman appears to address my question (see the theorem on p. 210) in that he proves (as you said) that almost all reals have infinitely many convergents with numerator and denominator the sum of two squares. $\endgroup$ – Victor Miller Aug 15 '13 at 23:35

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