Help with $\Bbb Z [\sqrt 5 ]=\{a+b\sqrt 5 | a,b \in \Bbb Z\}$ 
I've been working on this problem for a few hours now, and haven't been able to make much progress answering any of the parts. I'm looking more for a push in the right direction than direct answers. Any help is very much appreciated.  
These are my attempts at solutions  
a)
Let $A=a+b\sqrt d$, $B=x+y\sqrt d$  
$N(AB)=N(ax+\sqrt d (ay+bx+by \sqrt d))=(ax+\sqrt d (ay+bx+by \sqrt d))(ax-\sqrt d (ay+bx+by \sqrt d))$  
I tried to show that this is equivalent to this, but with no luck:
$N(A)N(B)=(a^2+db^2)(x^2+dy^2)=a^2x^2+da^2y^2+db^2x^2+d^2b^2y^2$
bi)
I can't completely wrap my head around the concept of units, so I was only able to use brute force to come up with units, and I have no way of knowing if I missed any. I tried using this equation, but didn't get anywhere.
$(a+b\sqrt{-5})(x+y\sqrt{-5})=1$
bii)
Similar problem to bi. I'm not sure how I might approach generating units.
 A: With your notation:
$$AB=(ax+byd)+(ay+bx)\sqrt d\implies N(AB)=(ax+byd)^2-(ay+bx)^2d=$$
$$=a^2x^2+\rlap{\;\;\;\;\;/}\color{green}{2abxyd}+b^2y^2d^2-a^2y^2d-\rlap{\;\;\;\;\;/}\color{green}{2abxyd}-b^2x^2d=$$
$$=a^2(x^2-y^2d)-b^2d(x^2-y^2d)=(a^2-b^2d)(x^2-y^2d)=N(A)N(B)$$
Now show that an element $\;w\in\Bbb Z[\sqrt d]\;$ is a unit iff $\;N(1)=\pm1\;$ ,so
$$w=a+b\sqrt{-5}\in\Bbb Z[\sqrt{-5}]\;\;\text{is a unit}\;\iff a^2+5b^2=\pm1\iff a=\pm 1\;,\;\;b=0 $$
Likewise, but really more interesting:
$$w=a+b\sqrt5\in\Bbb Z[\sqrt5]\;\;\text{is a unit}\;\iff a^2-5b^2=\pm1$$
For example, 
$$N\left(2+\sqrt{-5}\right)=4-5=-1\;\;\text{is a unit, and so are}\;\;\left(2+\sqrt{-5}\right)^n\;,\;\;n\in\Bbb N$$
A: Hint: In order to find units try to prove this intermediate statement: $u$ is a unit iff $N(u) = \pm 1$. Then try to solve the associated Pell's Equation. The problem is that from the Dirichlet Unit Theorem there are infinitely many solutions in the positive cases. But it also says that they are generated by one of them like powers. You don't need to prove that but just find you of them $\alpha$ and you got infinite of them $\alpha^n$ for every integer $n$.
