Diophantine equation (use class ideal group to solve) Use ideal class group to find all integer solutions to the equation $$x^3=y^2+200$$
My approach: Observe that $\mathbb{Z}[\sqrt-2]$ is the field of integers in the ring $\mathbb{Q}(\sqrt -2).$  Factorize the RHS of the equation to get $$(y-10\sqrt-2)(y+10\sqrt-2)$$ Thus we can consider the equation as an equality of $\textbf{principal ideals}$:$$(x^3)=(y-10\sqrt-2)(y+10\sqrt-2)$$
If we can show the principal ideals $(y-10\sqrt-2)$ and $(y+10\sqrt-2)$ are coprime, then by the unique factorisation of ideals in a Dedekind domain to to get $$(y+10\sqrt-2)=I^3$$ for some ideal $I$ in $\mathbb{Z}[\sqrt-2]$. In other words $I$ has order dividing $3$ in the ideal class group, moreover,by the order of the ideal class group of $\mathbb{Z}[-\sqrt2]$ is 1 to get $I$ is principal. 
Next  is to suppose $I=(a+b\sqrt -2)$ and get $(a+b\sqrt-2)^3=y+10\sqrt-2$ for some $a,b \in \mathbb{Z}$, then one can equating coefficients and solve for $y$...
But, how to show $(y-10\sqrt-2)$ and $(y+10\sqrt-2)$ are coprime ideals? I intend to prove by contradiction: Suppose they are not coprime, then there exists a proper prime ideal $P \subset \mathbb{Z}[\sqrt-2]$ such that $P$ contains both principal ideals, it follows that $$y-10\sqrt-2,   y+10\sqrt-2, 2y, 20\sqrt-2, x$$ are all in $P$, so the norm of $P$ divides the norm of each of the numbers above. Observe that $Norm(20\sqrt-2)=200=2^3*5^2$, if one can show the norm of $norm(x)=x^2$ is coprime to $200$ then we reach a contradiction.. But I can't do that in this case as $x$ may not be coprime to $200$? If so then how to show the two principal ideals are coprime?
 A: You may be able to push this kind of argument through: 
Suppose 5 divides both $y+10\sqrt{-2}$ and $y-10\sqrt{-2}$. Then $5\mid x^3$, so $5^3\mid x^3$, so $5^2$ must divide one or the other of $y\pm10\sqrt{-2}$, but neither of the numbers $(y\pm10\sqrt{-2})/25$ is in ${\bf Z}[\sqrt{-2}]$. Hence, 5 is not a common divisor. 
A: Whilst you can work with elements instead of ideals, as people have pointed out, you don't have to. I don't think it makes things appreciably simpler, especially if you're happy with using ideals. So how about this:
Notice that you don't need that the ideals are coprime, just that each is an exact cube.
Suppose a prime ideal $P$ divides both of $(y\pm10\sqrt{-2})$. Then it divides their sum, $(20\sqrt{-2}) = (5)(\sqrt{-2})^5$. Thus, either $P = (5)$ or $P = (\sqrt{-2})$. Notice that both of these ideals are their own conjugates. By conjugating the unique prime factorisation, the power $e$ of any prime ideal $Q$ dividing $(y+10\sqrt{-2})$ is the same as the power of $\bar{Q}$ dividing $(y-10\sqrt{-2})$, so $2e$ is divisible by $3$, hence so is $e$.
For completeness, note that we know $(5)$ is prime by applying Dedekind's criterion to $X^2+2$, which is irreducible in $\mathbb{F}_5$. We could also show that $5$ is prime in $\mathcal{O}_K$ (which you would have to do anyway using the other method, and (I think) requires a little more work, even if it is more elementary).
