Show that $3+2\sqrt2$ has infinite order in the multiplicative group of units $\Bbb{Z}[\sqrt2]$ 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 $\Bbb{Z}[\sqrt2]^\times$.
 A: 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.
A: 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 $1$.
A: 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 $n=1$.
