Verifying a subring of R? How do I verify that $\mathbb{Z}[\sqrt{2}] = \{ a +b\sqrt{2} \, | \,  a,b \in \mathbb{Z} \}$ is a subring of $\mathbb{R}$ ? I'm thinking that i have to show that it's a subgroup which is closed under multiplication, is that correct?

Also, I have to show that $\mathbb{Z}[\sqrt{2}]^*$ is infinite. It furthermore says I'm supposed to consider powers of $1+\sqrt{2}$ . I am, however, completely lost on this one. I thought $\mathbb{Z}^* = \{ -1,1 \}$, and therefore $\mathbb{Z}[\sqrt{2}]^*$ would be $={-\sqrt{2},-1,1,\sqrt{2}}$, but I'm obviously wrong.
 A: Given a ring $R$, a subset $S\subseteq R$ is a subring if it contains the multiplicative identity of R and is closed under the "subtraction" and multiplication of $R$.
For the second question, recall that $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}\,|\,a,b\in\mathbb{Z}\}$. So $\mathbb{Z}[\sqrt{2}]^*$ is the set of elements of the form $a+b\sqrt{2}$ ($a,b\in\mathbb{Z}$) that have multiplicative inverses. So think about the hint: does $x=1+\sqrt{2}$ have a multiplicative inverse? What about powers of $x$?
A: I'm going to try to answer your second question, since showing that it's a subring only needs you to verify the conditions in the definition of a subring.
It seems that you're having some trouble understanding what $\mathbb{Z}[\sqrt{2}]^*$ is. By definition $\mathbb{Z}[\sqrt{2}]^*$ is the group of units of the ring $\mathbb{Z}[\sqrt{2}]$. It consists of elements $\alpha = a + b \sqrt{2}$ with $a, b \in \mathbb{Z}$ such that $\alpha$ has a multiplicative inverse in $\mathbb{Z}[\sqrt{2}]$.
This just means that if $\alpha = a + b \sqrt{2}$ is in $\mathbb{Z}[\sqrt{2}]^*$, then there must exist another element $\beta = c + d\sqrt{2}$ in $\mathbb{Z}[\sqrt{2}]$ such that $\alpha \beta = 1$.
Another fact that might be useful is that since $\mathbb{Z}[\sqrt{2}]^*$ is a group, then given any two elements $\alpha_1, \alpha_2 \in \mathbb{Z}[\sqrt{2}]^*$, then their product $\alpha_1 \alpha_2 \in \mathbb{Z}[\sqrt{2}]^*$ also.
Certainly $\pm 1 \in \mathbb{Z}[\sqrt{2}]^*$, but $\pm \sqrt{2} \notin \mathbb{Z}[\sqrt{2}]^*$ because for $\pm \sqrt{2}$ to be in $\mathbb{Z}[\sqrt{2}]^*$, there must be an element $c + d\sqrt{2}$ with $c, d \in \mathbb{Z}$ such that $\pm \sqrt{2}(c + d\sqrt{2}) = 1$. But observe that
$$
\begin{align}
\pm \sqrt{2}(c + d\sqrt{2}) = 1 &\iff \pm c\sqrt{2} + \pm 2d = 1 \\
&\iff c = 0 \quad d = \pm\frac{1}{2}
\end{align}
$$
But then since $d = \pm \frac{1}{2} \notin \mathbb{Z}$, this shows that $\pm \sqrt{2} \notin \mathbb{Z}[\sqrt{2}]^*$. 
On the other hand, since $(1 + \sqrt{2})(-1 + \sqrt{2}) = 1$, then $1 + \sqrt{2} \in \mathbb{Z}[\sqrt{2}]^*$. 
Now, you have to show that $\mathbb{Z}[\sqrt{2}]^*$ has infinitely many elements, and for this you're provided with the hint to consider powers of $1 + \sqrt{2}$. Well, since we already know that $1 + \sqrt{2} \in \mathbb{Z}[\sqrt{2}]^*$ and that products of elements in $\mathbb{Z}[\sqrt{2}]^*$ also lie in $\mathbb{Z}[\sqrt{2}]^*$ then we can conclude that any power $(1 + \sqrt{2})^n \in \mathbb{Z}[\sqrt{2}]^*$. Finally, since $1 + \sqrt{2} > 1$ then each succesive positive power gets bigger, so that we have actually showed infinitely many different elements in $\mathbb{Z}[\sqrt{2}]^*$
$$
1 < 1 + \sqrt{2} < (1 + \sqrt{2})^2 < (1 + \sqrt{2})^3 < \cdots
$$
