When are these quotient rings isomorphic? 
$\mathbb{Z}[x]/(x^2-a) \cong \mathbb{Z}[x]/(x^2-b).$ For which
  $\mathbb{Z} \ni a, b$ it's true?

My first thought was: when $a = n^2, b = m^2$, where  $\mathbb{Z}\ni n, m.$ But i'm not sure that's an answer
 A: A pedestrian approach: let $a,b\in\mathbb{Z}$.
Assume that $\mathbb{Z}[x]/(x^2-a)\cong\mathbb{Z}[x]/(x^2-b)$. This means that there is a ring isomorphism,
$$\varphi:\mathbb{Z}[x]/(x^2-a)\to\mathbb{Z}[x]/(x^2-b).$$
I guess we're working with unital rings, so must have $\varphi(1)=1$. Also, there exists $p,q\in\mathbb{Z}$ such that $\varphi(x)=p+qx$. Then:
$$a=\varphi(a)=\varphi(x^2)=\varphi(x)^2=(p+qx)^2=(p^2+bq^2)+2pqx$$
from which we conclude:
$$\begin{cases}p^2+bq^2=a\\2pq=0.\end{cases}$$
We must have $q\neq0$ for otherwise $\varphi$ isn't surjective, hence from the second line, we must have $p=0$, and from the first line we conclude that $bq^2=a$.
Similarly, there exists $q'\in\mathbb{Z}$ such that $\varphi^{-1}(x)=q'x$ and $q'$ satisfies: $a(q')^2=b$.
Hence, $a=bq^2=a(qq')^2$ from which we conclude that: either $a=0$ (and then $b=0$) or $q^2(q')^2=1$ and hence $q^2=1$ and hence $a=b$.
Conclusion: the two are isomorphic if and only if $a=b$, in which case there are exactly two isomorphisms!
