Q:

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:

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 $n=1$.

