Finitely many $x,y \in \mathbb{Z}[i]$ such that $x^2=y^2+36$. How to prove that there only finitely many $x,y \in \mathbb{Z}[i]$ such that $x^2=y^2+36$.
Approach: I'm sorry but I've no idea how to start. I wrote $36=(x-y)(x+y)$, but I don't know if that's useful. Thanks in advance.
 A: Let $x=a+ib$ and $y=c+id$, where $a,b,c,d \in \mathbb{Z}$. Then the norm $N(x)=a^2+b^2$ and so on. The advantage of the norm is that it converts everything from $\mathbb{Z}[i]$ to $\mathbb{Z}$. One of the properties of the norm (can be easily proved) is that it is multiplicative, meaning $N(\alpha\beta)=N(\alpha)N(\beta)$.
\begin{align*}
36& = (x-y)(x+y)\\
N(36) & = N(x-y)N(x+y) &(\text{multiplicative property})\\
36^2 & =\left[(a-c)^2+(b-d)^2\right]\,\, \left[(a+c)^2+(b+d)^2\right]
\end{align*}
The last equation is in $\mathbb{Z}$, so we can use unique factorization idea to claim that the number of factors of $36^2$ is finite to claim that the number of $a,b,c,d \in \mathbb{Z}$ are finite.
If norm is something new to you, then:
We are given
$$36=(x-y)(x+y).$$
By taking conjugates, we get
$$36=\overline{(x-y)} \,\,\ \overline{(x+y)}.$$
Now multiply the two equations to get:
\begin{align*}
36^2 & = (x-y)(x+y) \,\, \overline{(x-y)} \,\,\ \overline{(x+y)}\\
36^2 & = (x-y)\, \overline{(x-y)} \,\, (x+y) \, \overline{(x+y)}\\
36^2 & =\left[(a-c)^2+(b-d)^2\right]\,\, \left[(a+c)^2+(b+d)^2\right]
\end{align*}
And we are back to what I had written above.
