In $\Bbb Z[\sqrt{2}]=\{a+b\sqrt{2}\rvert a,b∈\Bbb Z\}$, show that every element of the form $(3+2\sqrt{2})^n$ is a unit, where n is a positive integer

Question

In $\Bbb Z[\sqrt{2}]=\{a+b\sqrt{2}\rvert a,b∈\Bbb Z\}$, show that every element of the form $(3+2\sqrt{2})^n$ is a unit, where n is a positive integer.

My understanding of a unit is that if a is a unit, then ab = 1 = ba for some b. In other words, a is a unit if it has an multiplicative inverse.
Assuming I am correct, then I need to show that there is a multiplicative inverse for every $(3+2\sqrt{2})^n$.
Also, I am focusing on the last part of the question, or am I missing something from the definition at the beginning of the question?

If you show that $3+2\sqrt{2}$ is a unit, then all powers of it will be units. — Nishant, Apr 20 '15 at 00:40

score 8 · Accepted Answer · answered Apr 20 '15 at 00:41

8

Let us indulge in the hypothetical. $${1\over 3 + 2\sqrt{2}} = {3 - 2\sqrt{2}\over (3 + 2\sqrt{2})(3 - 2\sqrt{2})} = 3 - 2\sqrt{2}. $$ Et voila! It works. $3 + 2\sqrt{2}$ is a unit; thus all of its powers are, too.

answered Apr 20 '15 at 00:41

ncmathsadist

49,383

i.e. rationalize the denominator. (or consider the norm) – Bill Dubuque Apr 20 '15 at 00:43

score 2 · Answer 2 · answered Apr 20 '15 at 00:42

2

Your understanding is correct. Just don't fall into the pitfall right in front of you. If you can show that $3+2\sqrt 2$ is a unit, it immediately follows all powers of it are units too. So, you just need one verification, and the usual trick works.

answered Apr 20 '15 at 00:42

Ittay Weiss

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In $\Bbb Z[\sqrt{2}]=\{a+b\sqrt{2}\rvert a,b∈\Bbb Z\}$, show that every element of the form $(3+2\sqrt{2})^n$ is a unit, where n is a positive integer

2 Answers2