Showing that the only units in $\mathbb{Z}[i]$ are $1,\, -1, \, i, \, -i$? How would you show that the only units in $\mathbb{Z}[i] :=\{a + ib \, |\, a,b\in \mathbb{Z}\}$ are $1,\, -1, \, i, \, -i$?
 A: Hint: Every element of $\mathbb Z[i]$ has a natural number norm, and $\nu(xy)=\nu(x)\nu(y)$. This imposes a restriction on $\nu(x)$ if $x$ is a unit. 
A: The ring $\mathbb{Z}[i]$ is a subset of $\mathbb{C}$. Any nonzero $a + bi$ is invertible in $\mathbb{C}$, and the inverse is
$$(a + bi)^{-1} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i$$
Thus nonzero $a + bi \in \mathbb{Z}[i]$ is invertible in $\mathbb{Z}[i]$ if and only if $\frac{a}{a^2 + b^2} \in \mathbb{Z}$ and $\frac{b}{a^2 + b^2} \in \mathbb{Z}$.
Use this to show that $a + bi \in \mathbb{Z}[i]$ is invertible if and only if $a + bi = \pm1$ or $a + bi = \pm i$. To do this, treat the cases where $a = 0$ or $b = 0$ first. What happens when $a \neq 0$ and $b \neq 0$?
A: Here is an elementary proof that requires no knowledge of field theory (the field-theoretic proof requires not only the proof of multiplicativity of the norm, but also the proof of multiplicativity (and well-definedness!) of conjugation, a subtlety that is too frequently overlooked).
Let $\rm \ c = (a\!+\!b{\it i})(a\!-\!b{\it i}) = a^2 + b^2\in\mathbb Z.\:$ Then by rationalizing denominators we deduce
$$\rm \frac{1}{a\!+\!b{\it i}} = \frac{a\!-\!b\:\!{\it i}}{c}\in \mathbb Z[{\it i}\:]\:\Rightarrow\:\frac{a}c,\frac{b}c\in\mathbb Z\:\Rightarrow \frac{a\!+\!b{\it i}}c\in \mathbb Z[{\it i}\:]\:\Rightarrow\frac{a\!-\!b{\it i}}{c}\frac{a\!+\!b{\it i}}c = \frac{c}{c^2} = \frac{1}c\in\mathbb Z[{\it i}\:]$$
Thus $\rm\:1/c \in \mathbb Z[{\it i\:}]\cap\mathbb Q = \mathbb Z\:$ $\Rightarrow$ $\rm\: 1 = c = a^2 \!+\! b^2,\:$ so $\rm\:a = \pm 1, b = 0\:$ or $\rm\:a = 0, b = \pm 1.\ \ $ QED

Alternatively, the gcd $\rm\:\!(a,b)\! =\! d\:\!|\!\:a,\!b,\:$ so $\rm\:d^2\:\!|\!\:a^2,b^2\Rightarrow\:\color{#c00}{d^2\:\!|\:c}\!=\! a^2\!+\!b^2,\:$ so
$$\rm \frac{1}{a\!+\!b{\it i}} = \frac{a\!-\!b\:\!{\it i}}{c}\in \mathbb Z[{\it i}\:]\:\Rightarrow\:c \:|\:\color{#0a0}{a,b}\:\Rightarrow\: \color{#c00}{d^2\:\!|\:c}\:|\:(\color{#0a0}{a,b})\!=\!d\:\Rightarrow\:  d = 1 = c\qquad$$
A: $\mathbb{Z}[i]$ coincides with the lattice points in the complex plane. If you pick a lattice point outside the unit circle, its inverse lies inside. The only lattice point inside however is $0,$ which is not invertible. So any units must lie on the unit circle.
