Finding Infinite Units in \mathbb{Q}(\sqrt{21}) Using Continued Fractions

[math_grl]

How do you find an infinite number of units of \mathbb{Q}(\sqrt{21}) using the \sqrt(21)? I saw one example using continued fractions but do not know how to apply it in this case. I do have the periodic form of the continued fraction of \sqrt(21).

 

[disregardthat]

Q(sqrt(21)) is a field, every element except 0 is a unit. a/(a^2+b^2)-b*sqrt(21)/(a^2+b^2) is an inverse to a+b*sqrt(21) for rational a and b.

 

[Petek]

I interpret the original question to be "Find the units in the ring of integers of the number field \mathbb{Q}(\sqrt{21}). If so, then you may know that the units in that ring correspond to solutions of the so-called Pell's equation x^2 - 21y^2 = 1. Furthermore, solutions to Pell's equation can be obtained from the continued fraction expansion of \sqrt{21}.

If this is what you're looking for, then your textbook probably has more details and examples. I'd be glad to give more advice if you have specific questions.

 

[math_grl]

Yes Petek you hit the nail on the head, actually the book I'm reading seems to have no details and the few resources I looked up online just said we can derive some particular unit that if you raise it to the nth power, it's still a unit, from the pell's equation. I know the fundamental unit for \mathbb{Q}(\sqrt(21)) I'm just really unclear on the procedure or computation it involves to get there or to at least find a unit such as the one I described above.

 

[math_grl]

Thanks to Petek I was able to clear up my misunderstanding.