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Given an irrational algebraic number $\alpha$ (and maybe I want to add: of degree greater than $2$?), do there exist infinitely many relatively prime and square-free $p$,$q$ with $$|\alpha - p/q | < \frac{1}{q^2}\ .$$

I would guess "yes" based on a combination of Dirichlet's approximation theorem and the positive density of square-free integers among all the integers, but I can't even think of one example where I can prove this.

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Are there infinitely many pairs of consecutive squarefree fibonacci numbers? –  Daniel Litt Mar 25 '11 at 4:44
    
@Daniel I thought about that a little bit, and that motivated my parenthetical above. Experimentally, it seems most prime squares p^2 spoil about 2/p^2 of the consecutive Fibonacci pairs, so that would be good. But every now and then you come to a prime square such as $47^2, 89^2, 233^2$ that spoils pairs much more often. So it looks like the square-frees have a density in the Fibonacci's but not the same as the density they have among the natural numbers, and likewise square-free consecutives have a density, but I don't see how to get a proof! –  David Feldman Mar 25 '11 at 5:25
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1 Answer

I believe this is an open problem. Heath-Brown (1984) proved that there are infinitely many solutions to $|\alpha-p/q| < q^{-5/3+\epsilon}$ in square-free numbers. Harman (1984) proved that for almost all $\alpha$ there are infinitely many solutions to $|\alpha-p/q| < q^{-2+\epsilon}$ in square-free numbers. See paper1 and paper2.

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