Does there exist $a,b \in \mathbb{Z},$ $ a \neq b$, such $\mathbb{Z}[\sqrt{a}]$ $\cong$ $\mathbb{Z}[\sqrt{b}]$? Why? 
Does there exist $a,b \in \mathbb{Z},$  $ a \neq b$, and $\sqrt{a},\sqrt{b}\notin \mathbb{Z}$, such $\mathbb{Z}[\sqrt{a}]$ $\cong$ $\mathbb{Z}[\sqrt{b}]$? Why?

I know for sure, it is false that $\mathbb{Z}[\sqrt{-5}]$ $\cong$ $\mathbb{Z}[\sqrt{5}]$.
Also, how about the more general case that, for $p$ prime, does there exist $a,b \in \mathbb{Z},$  $ a \neq b$, such that $\mathbb{Z}[\sqrt[p]{a}]$ $\cong$ $\mathbb{Z}[\sqrt[p]{b}]$? 
$\phi\restriction_\mathbb{Z}$ $=$ $id$ and $\phi:\sqrt{5} $ $\mapsto$ $m+n*\sqrt{7}$,  for $m,n \in \mathbb{Z}$,and obviously $n \neq 0$, otherwise $\phi(\mathbb{Z}[\sqrt[p]{5}])=\mathbb{Z}$ , then $\phi $  is obviously not subjective.
$5=(\sqrt{5})^2,\phi(5)=\phi((\sqrt{5})^2)$=$\phi(\sqrt{5}))^2$ $= (m+n*\sqrt{7})^2 = m^2+7*n^2+2mn*\sqrt{7}$.
Therefore, $mn=0, m^2+7*n^2=7$, $25=5*5= ((m+n*\sqrt{7})^2)^2 = (m^2+7*n^2)^2$ $= m^4+49*n^4+14*(mn)^2 = m^4+49*n^4$.
Therefore $n=0$, contradiction.
 A: Isomorphic as what?
They are isomorphic as $\mathbb{Z}$ modules (i.e. Abelian groups).
They are not isomorphic as rings. The main point is that $\mathbb{Z}$ is uniquely included into any ring, so any ring isomorphism would have to preserve the integer part. From there, $\sqrt{a}$ must be mapped to $\sqrt{b}$ or $-\sqrt{b}$ (an isomorphism would have to preserve the set of generators of the subgroup of zero and the non-integer elements that square to an integer, if you like). Then it follows their squares must agree, so $a = b$.
A: *

*Since $A \cong B \implies \mathrm{Frac}(A) \cong \mathrm{Frac}(B)$ (the fraction field) if $\mathbb{Z}[\sqrt{a}] \cong \mathbb{Z}[\sqrt{b}]$ then $\mathbb{Q}(\sqrt{a}) \cong \mathbb{Q}(\sqrt{b})$.
But (since $\mathbb{Q}(\sqrt{a})/\mathbb{Q}$ is a Galois extension) the Galois theory tells us $\mathbb{Q}(\sqrt{a})\cong \mathbb{Q}(\sqrt{b})$ iff $\mathbb{Q}(\sqrt{a})= \mathbb{Q}(\sqrt{b})$, which means $a = c^2 b$, and the isomorphism must send $\sqrt{a}$ to $\pm \sqrt{a}$ (the roots of $x^2-a$).
Of course $\mathbb{Z}[\sqrt{a}] \not\cong \mathbb{Z}[c\sqrt{a}]$ since $\pm \sqrt{a} \not \in \mathbb{Z}[c\sqrt{a}]$ hence there are no such $a,b$.


*For $\sqrt[p]{a}$ it works the same way: with $\zeta_p$ a primitive root of unity,
if $\mathbb{Z}[\sqrt[p]{a}] \cong \mathbb{Z}[\sqrt[p]{b}]$ then $\mathbb{Z}[\sqrt[p]{a}][\zeta_p] \cong \mathbb{Z}[\sqrt[p]{b}][\zeta_p^k]$ (where $\zeta_p^k,p \nmid k$ is any other primitive root of unity)
and hence $\mathbb{Q}(\sqrt[p]{a},\zeta_p) \cong \mathbb{Q}(\sqrt[p]{b},\zeta_p^k)$.
But $\mathbb{Q}(\sqrt[p]{a},\zeta_p^k) /\mathbb{Q}$ is a Galois extension, so we must have $\mathbb{Q}(\sqrt[p]{a},\zeta_p) =\mathbb{Q}(\sqrt[p]{b},\zeta_p^k)$ and $\sqrt[p]{b} \in \mathbb{Q}(\sqrt[p]{a},\zeta_p)$ which means $b = c^p a$, and the isomorphism must send $\sqrt[p]{a}$ to $\pm \zeta_p^m\sqrt[p]{a}$ (the roots of $x^p-a$)
Of course $\mathbb{Z}[\sqrt[p]{a}] \not\cong \mathbb{Z}[c\sqrt[p]{a}]$ so there are no such $a,b$.
A: Yes, there are: $a=0$ and $b=1$. Or any pair of square integers.
To make the thing less trivial, let's assume $a$ and $b$ are distinct (nonzero) squarefree integers. It's not restrictive to assume $b\ne1$.
Suppose they are isomorphic. Then there are $x,y\in\mathbb{Z}$ such that
$$
(x+y\sqrt{b})^2=a
$$
so
$$
x^2+by^2+2xy\sqrt{b}=a
$$
Since $\sqrt{b}$ is irrational or $i$ times an irrational by assumption ($b\ne1$ is squarefree), we must have $x=0$ or $y=0$.
If $x=0$, then $a=by^2$ and, since $a$ is squarefree, $y^2=1$, so $a=b$: a contradiction.
If $y=0$, then $a=x^2$ and the only possibility is $a=1$. But then $\mathbb{Z}[\sqrt{b}]$ is isomorphic to $\mathbb{Z}$ as rings, which is impossible, because $\mathbb{Z}[\sqrt{b}]$ is a free abelian group of rank two.
