show that $(3+\sqrt{10})^n$ is a unit of $\mathbb{Z}[\sqrt{10}]$ for every $n \in \mathbb{Z}$ 
Show that $(3+\sqrt{10})^n$ is a unit of $\mathbb{Z}[\sqrt{10}]$ for every $n \in \mathbb{Z}$.

Should we prove that $\mathbb{Z}[\sqrt{10}]$ is a Euclidean domain first? Then what is the following proof?
 A: Hints: either of the following will work to show that $3+\sqrt{10}$ is a unit in $\mathbb{Z}[\sqrt{10}]\,$:


*

*look for integers $a,b \in \mathbb{Z}$ so that $(3+\sqrt{10})(a+b\sqrt{10})=1$ $\implies$ $3a+10b = 1$, $a+3b=0$;

*rationalize the denominator of $\cfrac{1}{3+\sqrt{10}}$ and show that the result belongs to $\mathbb{Z}[\sqrt{10}]$.
Once determined that $e = 3+\sqrt{10}$ is a unit in $\mathbb{Z}[\sqrt{10}]$ it follows that $e^n$ is a unit as well.
A: Define a norm on $\mathbb{Z}[\sqrt{5}]$ .
Take the function $N(a+b\sqrt{5})=(a+b\sqrt{5})(a-b\sqrt{5})=a^2-5b^2$
Prove that is is a norm i.e $N(0)=0$ and $N(xy)=N(x)N(y)$
Then prove that if $N(x)=1$ or $N(x)=-1$  then $x$ is a unit in $\mathbb{Z}[\sqrt{5}]$  and use all these properties of $N$ and induction on $n$ to prove that the element $(3+\sqrt{10})^n$  t is a unit.
$N(3+\sqrt{10})=-1$ hence...
A: Look up the definition of a unit in a ring.  Can you find an inverse for $3+\sqrt{10}$? If you can find an inverse for $3+\sqrt {10}$ what can you say about $(3+\sqrt{10})^n$?
A: We must have $(a+b\sqrt{10})(a-b\sqrt{10}) = +1$ or $- 1$ .The conjugate of $(3+\sqrt{10})$ is $(3-\sqrt{10})$ , and we have:
$(3+\sqrt{10}^)(3-\sqrt{10})= 9- 10 = -1$ so $3 + \sqrt{10}$ is a unit in $\mathbb{Z}[\sqrt{10}]$ for any $n \in \mathbb{Z}$.
See also : 'how to find units of in $\mathbb{Z}[\sqrt{n}]$ in this site' for more information.
