What is the multiplicative order of $1+\sqrt{2}$?

Actually I am in the context of Contemporary Algebra by Gallian, where there is topic of divisibility in integral domains, where there is inverse of $1+\sqrt{2}$ in $\mathbb Z[\sqrt{2}]$. I understand and know how to find inverse of $1+\sqrt{2}$, but this question after asked what is multiplicative order of $1+\sqrt{2}$? I want to know what is multiplicative order of $1+\sqrt{2}$?

I am not sure what is multiplicative order.

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By definition the multiplicative order of an element $x$ is the smallest positive integer $n$ (if it exists) such that $x^n=e$, where $e$ is the identity element of the set given. If no such $n$ exists then the order is said to be infinite.

In your case you should ask, is there a positive integer $n$ such that
$$(1+\sqrt{2})^n=1.$$

As you can see the left side is $>1$ so such a $n$ cannot exist. Thus the multiplicative order should be infinite.

The multiplicative order of $a=\sqrt{2}+1$ is $\infty$. This can be easily shown using the fact that the group of units of $\Bbb{Z}[\sqrt{2}]$ can be embedded (trivially by inclusion) in the multiplicative group $\Bbb{R}\setminus \{0\}$. So
$$a < a^2 < a^3 < \dots$$
as you can see, you can never have that $a^k=1$ for $k>0$. So all powers of $a$ are distinct.