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

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To find an infinite number of units in the number field \mathbb{Q}(\sqrt{21}), one can utilize the periodic continued fraction expansion of \sqrt{21}. The units in the ring of integers of this field correspond to solutions of Pell's equation x^2 - 21y^2 = 1. These solutions can be derived from the continued fraction expansion, which provides a systematic way to generate units. The discussion highlights the importance of understanding the fundamental unit and the process of raising it to the nth power to obtain additional units. Overall, the thread emphasizes the connection between continued fractions and the identification of units in quadratic fields.
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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).
 
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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.
 
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.
 
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.
 
Thanks to Petek I was able to clear up my misunderstanding.
 
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