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I would like some elaboration for the bound appearing in this answer.

Namely, let $u\in \mathbb Z[\sqrt 2]^\times $ be an invertible element of the ring $\mathbb Z[\sqrt 2]$ with $u>1$. Then, there exists some nonnegative integer $k\in \mathbb Z$ such that $$(1+\sqrt{2})^k\le u <(1+\sqrt{2})^{k+1}.$$ Do we use any Calculus argument here? Could you please give me a hand please?

user26857
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Chris
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    Consider the points $(1+\sqrt{2})^0, (1+\sqrt{2})^1, (1+\sqrt{2})^2, \dots$ on the number line -- the first point is just $1$, and the points increase without bound. Since $u > 1$, $u$ must lie between two adjacent points. That's all this is saying. – diracdeltafunk May 31 '21 at 23:55
  • For exactly the same reason, there exists a non-negative integer $\ell$ such that $3^\ell \leq u < 3^{\ell+1}$ -- every real number $\geq 1$ lies in one of these intervals! – diracdeltafunk May 31 '21 at 23:58
  • @diracdeltafunk Thanks for your comment. Ok, intuitively this seems to be fine. But how could we show this more formally? – Chris Jun 01 '21 at 00:00
  • I've posted a more formal argument as an answer; it's a very direct translation of the idea I wrote above into more mathematical language. Hope it helps! – diracdeltafunk Jun 01 '21 at 00:20

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Let $r = 1 + \sqrt{2}$ and let $u \in [1,\infty)$ be arbitrary. Since $r > 2$, we have $r^k \geq 2^k > k$ for all positive integers $k$, so the increasing sequence $r^0, r^1, \dots$ is unbounded. Thus, the set $\{n \in \mathbb{N} : r^n \leq u\}$ is bounded (i.e. finite) and nonempty (because $r^0 = 1 \leq u$). Let $N$ be the maximum element of this set. By maximality, $r^{N+1} > u$, so we have

$$(1+\sqrt{2})^N \leq u < (1+\sqrt{2})^{N+1},$$

as desired.

diracdeltafunk
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  • Very nice, thanks a lot! This was the type of answer I was looking for. – Chris Jun 01 '21 at 00:29
  • Why $u\in [1,\infty)$ and not $u\in(1,\infty)$? Also, should't it be $r^k>2^k>k$ since $r>2$ and $k\in \Bbb Z^+$? – Chris Jun 01 '21 at 13:42
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    (1) Just because $u \in [1,\infty)$ is a weaker assumption for which the same proof still works. (2) Indeed $r^k > 2^k$, but also $r^k \geq 2^k$ is true; it didn't matter which one we used for this proof. So, if you'd like, you can make both of those changes to the proof and it will still be entirely correct. – diracdeltafunk Jun 01 '21 at 17:39
  • (1) Yes, but if we take $u\in (1,\infty)$ in fact we exclude the case $u=1$. Thus, wouldn't it be better to take $u\in [1,\infty)$ as you suggested? (2) $r^k\geq 2^k$ is true for any nonnegative integer $k$, ie not positive integer. :) So we can wither say $r^k>2^k$ for any positive integer $k$ or $r^k\geq 2^k$ for any nonnegative integer $k$. Agree? :) – Chris Jun 01 '21 at 21:38
  • Agreed. These choices don't matter for the validity of the proof of the desired claim. I wrote what felt stylistically best to me, and anyone is welcome to agree or disagree with those stylistic choices! – diracdeltafunk Jun 01 '21 at 21:48
  • Yes, and this is absolutely fine, but I mention this as you write "$r^k\geq 2^k>k$ for all positive integers $k$" instead of "for all nonnegative integers $k$". – Chris Jun 01 '21 at 21:52