Prove that only units of $\Bbb Z[\sqrt d] $ are $\pm 1$. 
Let $d(\in \Bbb Z)<-1$ such that $d$ is not divisible  by the square of a prime.
  Prove that only units of $\Bbb Z[\sqrt d] $  are $\pm 1$.

$$a+b\sqrt d \text{ is a unit }\implies (a+b\sqrt d)(c+e\sqrt d)=1\implies (a^2-db^2)(c^2-de^2)=1\implies a^2-db^2=\pm 1\implies a+b\sqrt d=\pm1\implies a+b\sqrt d \text{ is a unit}$$    
Where did we use that $d$ is not divisible  by the square of a prime.
Is my proof wrong?
 A: This is why some people prefer to use $-d$ with $d > 0$ rather than $d < 0$. With $-d$, you can then write the norm of a number as $a^2 + db^2$. It then becomes obvious that you can't have $a^2 + db^2 = -1$. Furthermore, as $a$ and $b$ get further away from 0, the norm is correspondingly larger.
So whatever units there are in the ring must have small $a$ and $b$ and therefore be quite close to 0. Thus, for example, in $\mathbb Z[\sqrt{-5}]$, we see that $-1$ has a norm of 1 and is therefore a unit, but already $-1 + \sqrt{-5}$ has a norm of 6 and indeed any number in this ring with both $|a|$ and $|b|$ greater than 1 must have a norm greater than 6.
Oh, but we have to mind the so-called "half-integers". For example, $$N\left(-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = 1.$$ However, $$N\left(-\frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = \frac{1}{4} + \frac{7}{4} = 2,$$ so that does it for half-integers that are units.
This is not a rigorous proof, but I do think there is enough here for you to make it one.
A: I'm going to assume that you meant $d \not \equiv 1 \bmod 4$. If you did mean that, my answer might need slight adjustments.

$$a + b\sqrt d \text{ is a unit }\implies (a + b\sqrt d)(c + e\sqrt d) = 1$$

So far so good. Of course it should be noted that $\{a, b, c, e\} \in \mathbb{Z}[\sqrt{d}]$. More importantly, however, either $c = -a$ and $e = b$ or $c = a$ and $b = -e$. Without loss of generality (I'm using that correctly, right?), let's rewrite $c + e \sqrt{d}$ as $a - b \sqrt{d}$. Then

$$(a + b\sqrt d)(a - b \sqrt d) = 1 \implies (a^2 - db^2)(a^2 + db^2) = 1 \implies a^2 - db^2 = \pm 1$$

It is at this point that I would invoke Mr. Soupe's shortcut. Replace $d$ by $-d$ and replace $a^2 - db^2$ with $a^2 + db^2$. Then $d$ is positive and $a^2 + db^2 = -1$ has no solutions. If $d > 1$ (remember the replacement we made just a minute ago), and $b \neq 0$, then $db^2 > 1$ and $a + b \sqrt{d}$ can't be a unit.
So for that number to be a unit, we require $b = 0$. And then if $a > 1$ or $a < -1$ then $a^2 > 1$. I don't have to tell you what happens if $a = 0$. Then, by the process of elimination, $a = \pm 1$, $b = 0$ and therefore the only numbers in the group of units are $\{-1, 1\}$.
A: You're begging the question. You prove that $a^2-db^2=\pm1$, but you can't conclude right now that $a+b\sqrt{d}=\pm1$. What you can derive is that $a+b\sqrt{d}$ is a unit with inverse $\pm(a-b\sqrt{d})$.
Actually, since $d<0$, $a^2-db^2\ge0$, so you can only have $a^2-db^2=1$.
Now $a^2-db^2\ge a^2$. If this is to be equal to $1$, we need $a=0$ or $a=\pm1$. In the first case $-db^2=1$ is impossible, as $d<-1$. So $a=\pm1$ and therefore $b=0$.
There is no need to assume that $d$ is squarefree. This assumption is relevant when you want to consider $\mathbb{Q}[d]$, because this is the same as $\mathbb{Q}[d']$ where $d'$ is squarefree and $d=x^2d'$ for some integer $x$. The field $\mathbb{Q}[d]$ is obviously related to $\mathbb{Z}[d]$, being its field of quotients.
