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?
Best Answer
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.