0

Consider the system $\mathscr A$ of numbers of the form $c+d\sqrt 2$ where $c,d\in \mathbb Z$. If $\alpha, \beta \in \mathscr A$, we define divisibility by $$\alpha|\beta :\equiv \exists \gamma\in \mathscr A \space \space \text{s.t} \space \space \alpha \gamma= \beta.$$ We call $\mu$ a unit if it divides all the numbers in the system. The problem is to show that all the units are of the form $$\pm (1+\sqrt 2)^k.$$ What I know is that $c+d\sqrt {2}$ is a unit if and only if $|c-2d^2|=1$. Let’s assume by contradiction that there is a unit $a+b\sqrt2$ of a different form. The problem is to find a suitable choice of a unit of the above-mentioned form so that their product $u+v\sqrt 2$ will satisfy $1\lt u+v\sqrt 2\lt 1+\sqrt 2$, in contradiction with the above-mentioned modulus characterization.

My problem is that I can’t find such a suitable factor to reach the required inequality. How would we do so?

  • 2
    If $\alpha$ is a unit, then so $\alpha (\sqrt2 -1)$. Therefore, if $\alpha >1$, then $\alpha (\sqrt2 -1)^n\le1$ for some $n$. – lhf Feb 23 '22 at 11:52
  • @lhf I got it now, thanks. However, I am dubious about why you considered only the case $\alpha\gt 1$. Is it to reduce it to the case $\beta \lt 1$ so that the required inequality holds? –  Feb 23 '22 at 13:48
  • I don't understand what you are trying to do. $1+\sqrt2$ is the fundamental unit because a unit $u$ is a root of $x^2-ax\pm 1$ with $a\in \Bbb{Z},a=u\pm 1/u$. If $u \in (1, 1+\sqrt2)$ then $u\pm 1/u \in (0, 3)$, not many choices to check. – reuns Feb 23 '22 at 13:59
  • @reuns I’ll write up an answer to make it more clear –  Feb 23 '22 at 15:12
  • @user26857 I wanted to solve this problem with the method outlined in my question. However, thanks for your suggestion containing different proofs –  Feb 23 '22 at 16:23

1 Answers1

0

If $\alpha$ is a (positive) unit that is not of the above-mentioned general form, then we proceed as follows.

Case 1 ($\alpha \lt 1$). We multiply it by $(1+\sqrt 2)$ for a sufficient number of times so that $1\lt u+v\sqrt 2:= \alpha(1+\sqrt2)^k\lt (1+\sqrt2)$.

Case 2 ($\alpha \gt 1$). Similarly, we multiply it with negative powers of $(1+\sqrt 2)$ so that we obtain the same inequality.

In both cases, $u$ and $v$ can’t be both positive as it would contradict the right inequality. They can’t be both negative as well as it would contradict the left inequality . Obviously the zero value would be contradictory too.

So let’s assume $v$ is negative, so $u$ will be positive. Multiplying the left inequality by $u-v\sqrt 2$ we will get $u-v\sqrt 2 \lt +1$, so we will have $u\lt 1+v\sqrt 2\lt 1$, contradicting the positivity of $u$. Similarly for the other case.