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So as a beginning algebraic number theory enthusiast, my question is about the integer solutions of certain equations (let's say Euler's solution to $y^2 = x^3 - 2$). We work over $\mathbb{Z}[\sqrt{-2}]$ as that is a UFD. I understand the way we get to the solutions $(3, \pm 5)$.

However, my question is about how does the solvability in $\mathbb{Z}[\sqrt{-2}]$ guarantee these same solutions in $\mathbb{Z}$? Is it because $\mathbb{Z} \subset \mathbb{Z}[\sqrt{-2}]$, and so solvability with solutions in $\mathbb{Z}[\sqrt{-2}]$ guarantees solutions in $\mathbb{Z}$ with the same solutions?

konjecture
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    Yes it's because $\mathbb Z\subset \mathbb Z[\sqrt{-2}]$. At the end of the day, you've found some integer solutions. The fact that you had to look in a larger ring to find them doesn't change the fact that they're integer solutions. (In the same way, the algorithm to solve cubic polynomials uses complex numbers, even when the polynomial has only real solutions) – Mathmo123 Nov 08 '21 at 06:49
  • Related: If you want to find integer roots of a polynomial, you use the rational roots theorem, then only pay attention to the roots you found that happen to be integers... – Eric Towers Nov 08 '21 at 06:50
  • @Mathmo123 - Thanks for the clarification. My follow up question now is, yes we find the integer solutions to the original equation by looking at the ring $\mathbb{Z}[\sqrt{-2}]$, however, what is the guarantee that there is no other "hidden" integer solutions of the equation in a different ring of integers? I mean we didn't check all possible rings of integers for solutions? Sorry, if it sounds like a very trivial question. – konjecture Nov 08 '21 at 07:02
  • Surely there are other solutions in other rings. Like $x=1, y=i$ in the ring $\Bbb{Z}[i]$ of Gaussian integers. But why would that be relevant to the original problem? The point was to use the properies of a bigger ring to deduce facts about possible solutions in $\Bbb{Z}$. The choice of a bigger ring that helps in a given situation is kind of a black art. – Jyrki Lahtonen Nov 08 '21 at 07:27
  • @JyrkiLahtonen Correct, we are only finding solutions in $\mathbb{Z}$ of the original equation. For that we looked at $\mathbb{Z}[\sqrt{-2}]$ as it was the natural choice since $y^2 + 2$ factors as $(y + \sqrt{2})(y - \sqrt{2})$. I was just curious for example - what is the guarantee that we will not find any $\mathbb{Z}$ solutions when looking at let's say the ring of integers of $\mathbb{Q}(\sqrt{-3})$? Or, in other words, how can we say that we found ALL the possible integer solutions? – konjecture Nov 08 '21 at 07:33
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    If we find all the solutions in some ring $R\supset \mathbb Z$, then we've necessarily found all the integer solutions too. So if we find all the solutions in $\mathbb Q(\sqrt{-3})$, then we'll have found all the integer solutions. That's all great abstractly, but on a practical level, that doesn't mean that all choices of $R$ are equal: if we can find all the solutions in $R$, then great, but some choices of $R$ are much more suitable to finding solutions than others. – Mathmo123 Nov 08 '21 at 07:44
  • Okay, thanks for all the answers. That makes more sense now. – konjecture Nov 08 '21 at 08:12
  • See also motivation for the notion of algebraic integer. – Bill Dubuque Nov 08 '21 at 08:51

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