A (partial) answer is given here: http://www.cs.ubc.ca/~gyomalin/docs/Thesis_GuillaumeAlain_2006juillet29.pdf. In it he shows that an irrational real can be simultaneously approximated well by infinitely many algebraic number of degree n (for fixed n). On the other hand, on page 39 (section 5) he looks at the case of a fixed quadratic imaginary field, and shows that there can't be an infinite number of good approximations. Victor On Wed, Aug 14, 2013 at 11:04 AM, Dan Asimov <dasimov@earthlink.net> wrote:
This is a very interesting question. We know that the rational points on the unit circle are all of the form (a/c, b/c) where (a,b,c) is a Pythagorean triple. And we know these are dense in the unit circle, so this approximation must be possible. (In fact these rational points form an interesting subgroup of the unit circle group SO(2).)
--Dan
On 2013-08-14, at 10:23 AM, Henry Baker wrote:
Suppose I'm trying to compute some rational approximation to a real number x, but with some additional constraint.
In particular, if m/n approximates x, I'd like sqrt(m^2+n^2) to be integral.
Suppose I used a continued fraction process to generate better & better approximations.
Is there any reason to believe that I'd eventually find one m/n for which sqrt(m^2+n^2) is integral ?
I did a quick search on both pi and e, and so far _none_ of the rational approximations (except for early integral approximations) m/n has sqrt(m^2+n^2) integral.
So this leads me to believe that perhaps what I'm trying to do is impossible; perhaps sqrt(m^2+n^2)=integer _never_ happens for rational approximations produced by continued fractions?
Or perhaps this situation is exceedingly rare.
If so, how rare is it?
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