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I had a question for the first two parts of this question here. I was told to post a new question for the last two parts. I have a hint concerning the question, but i'm not sure how it relates. Here's the hint:

If $a^n=1$, then $|a^n|=1$. If $0<a<1$, how does $|a^n|$ relate to $|a|$ for a positive integer n? If $|a|>1$, how does $|a^n|$ relate to $|a|$ for a positive integer n?

I understand that $a^n<a$ for $0<a<1$ and that $a^n>a$ for $a>1$, but I don't see how that relates to the question.

The last part of the question is: Show that -1 and 1 are the only units of $\Bbb{Z}[\sqrt2]$ that have finite order in $\Bbb{Z}[\sqrt2]^\times$.

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user21
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  • Surely $|a|$ denotes the order of $a$, not the absolute value of $a$, right? So the hint is asking how the order of $a^n$ is related to the order of $a$. – MPW Feb 24 '17 at 15:30
  • Wow. I didn't even think of that. I'm confusing my real analysis class with my modern algebra class now lol – user21 Feb 24 '17 at 15:31
  • :) ${}{}{}{}{}{}{}{}{}{}$ – MPW Feb 24 '17 at 15:33
  • Okay, so I know that $3+2\sqrt2$ is a unit in $(\Bbb{Z}[2])$. I also know that $3^2-2*(2^2)=1$. I'm stuck there though. I'm guessing that multipying $3+2\sqrt2$ by itself $n$ amount of times will never produce 1, so the order is infinite? – user21 Feb 24 '17 at 15:34
  • @johnie4usc yes, that's the basic idea behind showing something is not of finite order (i.e. following the definitions) – Adam Hughes Feb 24 '17 at 15:41
  • Okay. I'm just not sure how to format that in a proof type manner since I'm basically assuming that $(3+2\sqrt2)^n=1$ has no solution. – user21 Feb 24 '17 at 15:42
  • Yes, that's essentially it. You really want to show that no power of that number is ever $1$, but that's equivalent (because multiplying one more time would then give the original number again as you say). It looks like there is an answer posted below demonstrating how to show this. (Remember, the order of $a$ is the smallest positive integer $n$ such that $a^n=1$.) – MPW Feb 24 '17 at 15:43
  • Yeah, we've just never used summation, representations, or the binomial theorem in my class so I feel like it would be out of left field to bring that up. – user21 Feb 24 '17 at 15:44
  • "Modern algebra" was modern during the 1930s. It is safe to stop calling it that now... – Mariano Suárez-Álvarez Feb 24 '17 at 15:47
  • That's fine. Just the name of the class lol – user21 Feb 24 '17 at 15:48
  • You could just observe that all positive powers of $3+2\sqrt{2}$ are of the form $a+b\sqrt{2}$ with $a\geq 0$ and $b\geq 0$; multiplying one more time gives $$(3+2\sqrt{2})(a+b\sqrt{2})=(3a+4b)+(2a+3b)\sqrt{2}$$ and the integer part $3a+4b$ is larger than $a$ (the integer part from the previous power). This means the integer part is strictly increasing as the power increases, so you will never end up with $1+0\sqrt{2}$. – MPW Feb 24 '17 at 15:50
  • That makes sense. Thank you. – user21 Feb 24 '17 at 15:52
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    @MPW Why do you think $|a|$ denotes the order instead of absolute value? – Bill Dubuque Feb 24 '17 at 15:55
  • @BillDubuque : That was just my take on it (it's the usual notation in group theory). I suppose either interpretation gives a legitimate hint. Perhaps absolute value gives a better hint, since if $|a|\neq1$ then you have a sequence which diverges monotonically away from $1$. +1 for you. – MPW Feb 24 '17 at 16:02
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    @MPW But the hint doesn't seem to make any sense using order. How did you view it like that? – Bill Dubuque Feb 24 '17 at 16:17
  • @BillDubuque : Okay, you've squeezed it out of me. Upon further scrutiny, I see that my interpretation of the hint is completely wrong (apologies to johnie4usc). I retract my original comment (but I will leave it there since the subsequent responses won't make any sense to anyone reading them if the comment disappears). Another +1 for you. – MPW Feb 24 '17 at 16:24
  • @johnie4usc you don't assume it doesn't, you assume it does and derive a contradiction. See the structure of my answer to see how I establish that. – Adam Hughes Feb 24 '17 at 16:40
  • @johnie4usc in particular, it seems you understand the nature of MPW's suggestion, that is the essence of my answer as well, except I do the general case with explicit numbers instead of $a$ and $b$. – Adam Hughes Feb 24 '17 at 16:51
  • Yeah I understand it. My professor did use absolute value when he showed us the answer. – user21 Feb 24 '17 at 16:52
  • @AdamHughes Does your prior comment imply that you have some way to interpret the hint sensibly in terms of order vs. absolute value? If so please explain how you can read it that way. – Bill Dubuque Feb 24 '17 at 17:03
  • @BillDubuque no, sadly: I opted to go for what I felt was a much more straightforward route than following the hint. From experience, I'm sure $|\cdot|$ is intended to be the ordinary absolute value on $\Bbb R$--that is the standard approach. It reads that way exactly because of the $|a|>1\implies |a|^n\to\infty$, so the topology gives the result pretty easily. – Adam Hughes Feb 24 '17 at 17:11
  • @johnie4usc you should come back with something in the flavor from the answers you've gotten for this question, show him how you can approach the problem in multiple ways! :) – Adam Hughes Feb 24 '17 at 17:12
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    To use algebraic number theory terminology: the fundamental unit $\eta$ of this domain has a norm of $-1$, whereas $N(3 + 2 \sqrt 2) = 1$. Observe that $\eta^2 = 3 + 2 \sqrt 2$. Then $N(\eta^n) = \pm 1$ according to the parity of $n$. – Robert Soupe Feb 24 '17 at 19:12

3 Answers3

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Simply show that in the representation of $(3+2\sqrt{2})^n=a+b\sqrt{2}$ that $a>1$ always. But this is easy, use the binomial theorem to see

$$(3+\sqrt{2})^n=\sum_{k=0}^n{n\choose k}2^{k/2}3^{n-k}$$

Then we have

$$a=\sum_{k=0}^{n/2}{n\choose 2k}2^k3^{n-2k}>3.$$

So it is not of finite order.

To see the last part of your question, note that $m^2-2n^2=1$ of finite order means

$$(m+n\sqrt{2})^k=1$$

and again, if $m,n>0$ then we can appeal to the same argument, as we can if $m,n<0$. So we need that $mn<0$ or $mn=0$. Clearly if $mn=0$ we have $m=\pm 1$, so that case is easy, if $mn<0$ then WLOG assume $m>0$ and note that $m-n\sqrt{2}=(m+n\sqrt{2})^{-1}$ so $m+n\sqrt{2}$ has finite order iff $m-n\sqrt{2}$ does, and this is again impossible since here with $n'=-n$ we would need to show that $m+n'\sqrt{2}$ has finite order with $m,n>0$ a contradiction.

Adam Hughes
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    Even easier: the (real) absolute value of the powers of $3+2\sqrt{2}$ grows without bound, so can never equal $1$. – Ethan Bolker Feb 24 '17 at 15:38
  • @EthanBolker there are definitely alternatives, but I think that comment more belongs on the op's post than mine. :) – Adam Hughes Feb 24 '17 at 15:39
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I interpret your condition on the finite order $n$ of $a$ as $a^n = 1$ . But then your problem becomes simply a field theoretic one : the field $\mathbf Q (\sqrt 2)$ being embeddable into $\mathbf R$, the only roots of unity that it contains are $\pm 1$. This answers at the same time to both your questions. Note that your integral element $3+2 \sqrt 2 $ is a unit of $\mathbf Z [\sqrt2]$ because it has norm $1$.

nguyen quang do
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Since $$ \frac{1}{3+2\sqrt{2}}=3-2\sqrt{2} $$ if $(3+2\sqrt{2})^n=1$, then also $(3-2\sqrt{2})^n=1$. Suppose $$ (3+2\sqrt{2})^n=(3-2\sqrt{2})^n $$ Then $$ \left(\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\right)^n=1 $$ that is $$ (17-12\sqrt{2})^n=1 $$ Hence $$ \sum_{0\le k\le n}\binom{n}{k}(-1)^k17^{n-k}\cdot12^k\cdot(\sqrt{2})^k=1 $$ and so $$ -1+\sum_{\substack{0\le k\le n\\k\text{ even}}} \binom{n}{k}17^{n-k}\cdot12^k\cdot(\sqrt{2})^k= \sum_{\substack{0\le k\le n\\k\text{ odd}}} \binom{n}{k}17^{n-k}\cdot12^k\cdot(\sqrt{2})^k $$ The left-hand side is rational, the right-hand side isn't unless $n=1$.

egreg
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