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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?

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Why do you have to use the ideal class group, it is not needed here since the class number is 1...you are in a UFD! –  fretty Oct 12 '12 at 8:38

2 Answers 2

up vote 2 down vote accepted

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.

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How does this imply the principal ideals $(y+10\sqrt-2)$ and $(y-10\sqrt-2)$ are coprime? –  user31899 Oct 12 '12 at 12:22
    
First of all, stop talking about ideals - as fretty notes, you're in a UFD, you can deal with elements instead of ideals. Second, what I have showed you is that 5 can't be a common divisor. It's up to you to see if you can understand my argument and then adapt it to rule out other common divisors. You already know any common divisor must divide $20\sqrt{-2}$, so there aren't that many to try. In short, I haven't solved the problem for you --- I have shown you a possible way forward. Can you take up where I left off? –  Gerry Myerson Oct 12 '12 at 12:28
    
I see what you mean now. Similary $2$ and $\sqrt -2$ are not common factors, so gcd$(y+10\sqrt-2,y-10\sqrt-2)=\pm 1$ implies the ideals $(y+10\sqrt-2)$ and $(y-10\sqrt-2)$ are coprime due to we are in a UFD. The rest of my above arguments becomes valid then. Thanks! –  user31899 Oct 12 '12 at 13:19

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).

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