In order to study the equation $x^2 + y^2 = n$ I'd like to understand the related commutative ring $$ R = \mathbb{Z}[x,y]/(x^2+y^2-n)$$ I'd like compute the spectrum of this ring. What are the prime ideals?
$(x-a, y-b)$ with $a^2+b^2=n$
if $p(x,y)$ is an irreducible polynomial in $\mathbb{Z}[x,y]$ that is not a multiple of $x^2 + y^2 = n$, then $(f(x,y))$ is prime in $\mathbb{Z}[x,y]/(x^2+y^2-n)$. This is like considering $(f(x,y), x^2+y^2-n)$ inside of $\mathbb{Z}[x,y]$.
Are there any others? These polynomials may correspond to generic points (also here)?
Certainly if $\mathbb{Z}$ is an integral domain, so is $\mathbb{Z}[x,y]$ and since $x^2 + y^2 = n$ is irreducible, so is $\mathbb{Z}[x,y]/(x^2 + y^2 = n)$. This tells a few things, such as $(0)$ is prime.
it is not a PID e.g. $(2,x)$ is still prime.
Could there be a Gröbner basis approach. Since $\mathbb{Z}[x,y]$ is generated by monomials $x^a\, y^b$ maybe we could look at the images mod $x^2 + y^2 - n$. In that case:
$$ x^a \,y^{2b+(0\text{ or }1)} \equiv x^a \, (n - x^2)^b \, y^{(0\text{ or }1)} $$
At least, this would qualify what I mean by the word "compute". The primes over $\mathbb{Z}[x,y]$ are the irreducibles and we could ask whether any of these split over $(x^2 + y^2 - n)$.
The ring $R=\mathbb Z[x,y]:= \mathbb{Z}[X,Y]/\langle X^2+Y^2-n\rangle $ has Krull dimension $2$, so is certainly not a PID since PID's have Krull dimension $1$.
Let's now have a look at the scheme $S:=\operatorname {Spec}R$, which is an "arithmetic surface".
$\bullet$ The ring $R$ has exactly one prime ideal of height zero: the zero ideal $\langle 0_R\rangle \subset R$.
It corresponds to the generic point $\eta$ of the scheme $S$.
$\bullet \bullet$ The ring $R$ has prime ideals of height $1$: such an ideal may be thought of as the generic point of an integral curve drawn on the surface $S$.
For example the prime ideal you mentioned, $\langle x-a,y-b \rangle$ where $a^2+b^2=n$, has height one.
But there might be other prime ideals of height one: for example if $n=3$ we may take the ideal $\langle x\rangle$.
$\bullet \bullet \bullet$ Finally we have the prime ideals of height two, which are exactly the maximal ideals of $R$.
They correspond to the closed points of our surface $S$.
Supposing again that $n=3$, the prime ideal $\langle x,y\rangle$ is such a maximal ideal.