How to show that if $\alpha \in \mathbb{Z}[\sqrt{2}]$ and $\alpha$ is a unit, then we cannot have $1 < \alpha < 1 + \sqrt{2}$. How can I show that if $\alpha \in \mathbb{Z}[\sqrt{2}]$ and $\alpha$ is a unit, then we cannot have $1 < \alpha < 1 + \sqrt{2}$.
Assuming $\alpha = a + b\sqrt{2}$ is a unit, $1 < \alpha < 1 + \sqrt{2}$.
Also considering $\beta = a - b \sqrt{2}, \alpha \cdot \beta = \pm1$ and an absolute value estimate of $\beta$...
 A: Let $z=a+b\sqrt{2} > 1$ be a unit. This gives rise to 4 units $\pm z, \pm \frac{1}{z}$. In terms of $a$ and $b$ they are written as $\pm a \pm b\sqrt{2}$. Only one of those $4$ is greater than $1$. From this we can deduce $a,b>0$, because the greatest number of the $4$ numbers $\pm a \pm b\sqrt{2}$ is the one, where both coefficients occur positive. This shows the assertion.
A: Basically you're being asked to prove that in between $1$ and $1 + \sqrt{2}$ (which are both units in $\mathbb{Z}[\sqrt{2}]$, by the way) there are no other units. The question may seem daunting because there are infinitely many units in $\mathbb{Z}[\sqrt{2}]$.
But those units follow a regular, orderly procession. A unit in this domain must be of the form $(\pm 1 \pm \sqrt{2})^n$, with $n \in \mathbb{Z}$. We have four cases to consider:


*

*$(1 + \sqrt{2})^n$: As $n$ gets larger, so does $(1 + \sqrt{2})^n$. It should be enough to check $(1 + \sqrt{2})^n$ for $-5 < n < 5$. Any greater value of $n$ will make $\alpha = (1 + \sqrt{2})^n$ too distant from $1$ and $1 + \sqrt{2}$ to possibly fall in the narrow range $1 < \alpha < 1 + \sqrt{2}$.

*$(1 - \sqrt{2})^n$: As $n$ approaches $-\infty$, the absolute value of $(1 - \sqrt{2})^n$ gets larger, while as $n$ approaches $+\infty$, the absolute value of $(1 - \sqrt{2})^n$ gets closer to $0$.


If you absolutely want to be sure, also check $(-1 + \sqrt{2})^n$ and $(-1 - \sqrt{2})^n$. You won't be surprised by the results.
