# Solving a Diophantine Equation using factorisation of ideals

I am stuck on the following question which is given as follows:

Prove that the only integer solutions to the equation $$x^2 + 13 = y^3$$ are $(70,17)$ and $(-70, 17)$.

(Hint: first show that $x$ is even and $y$ is odd)

I have seen a solution to this and first part to it is as follows.

Let $x$ and $y$ satisfy the given equation. A odd square is $1$ modulo $4$ and a cube cannot be $2$ modulo $4$ so $x$ is even, which immediately implies $y$ is odd.

So we have a factorisation of ideals $(x + \sqrt{-13}) \cdot (x - \sqrt{-13}) = (y)^3. \$ The gcd of $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$ divides $(2\sqrt{-13}) = (2, 1+\sqrt{-13})^2 \, (\sqrt{-13}). \$ As $y$ is odd, $(y)$ cannot be divisible by $(2, 1+\sqrt{-13})$ hence $$(x + \sqrt{-13}, \, x - \sqrt{-13}) = (1) \hspace{5pt} \text{or} \hspace{5pt} (\sqrt{-13}) \hspace{5pt} (*)$$ In the latter case, $x$ would be divisible by $13$, hence $y$ would be divisible by $13$ but then $13 \ \big| \ y^3$ and $x^2 + 13 \equiv 13 \pmod{13^2}$ which is a contradiction so $(x + \sqrt{-13}, \, x - \sqrt{-13}) = (1)$. It follows that the principal ideal $(x + \sqrt{-13})$ is the cube of an ideal $I$.

So firstly why does $(*)$ hold given that $(y)$ cannot be divisible by $(2, 1 +\sqrt{-13})$ and in the latter case why does this mean $x$ is divisible by $13$?. Finally how do we know that $(x + \sqrt{-13})$ is the cube of an ideal $I$?

Any help would be very much appreciated!

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For your first question: Let $P$ denote the gcd of $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$. You have shown that $P$ divides $(2,1 + \sqrt{-13})^2(\sqrt{-13})$. But also $$P\mid (x - \sqrt{-13})(x + \sqrt{-13}) = (y)^3.$$ Since $P$ divides both $(y)^3$ and $(2,1 + \sqrt{-13})^2(\sqrt{-13})$, it must divide their gcd. You know that $(2,1 + \sqrt{-13})\nmid (y)$, and hence $(2,1+\sqrt{-13})\nmid (y)^3$, so the gcd of $(2,1+\sqrt{-13})^2(\sqrt{-13})$ and $(y)^3$ divides $(\sqrt{-13})$. Therefore $P\mid (\sqrt{-13})$. I hope that clarifies the first point.
For the second question: assuming $P = (\sqrt{-13})$. Since $P$ divides both $(x - \sqrt{-13})$ and $(x + \sqrt{-13})$, you get $$(13) = P^2 \mid (x + \sqrt{-13})(x - \sqrt{-13}) = (x^2 + 13),$$ so $13\mid x^2 + 13$. It follows that $13\mid x$.
For the third question. We assume that $P = (1)$, and want to understand why $(x + \sqrt{-13})$ is the cube of an ideal. This will follow from the starting equation $$\tag{**}(x -\sqrt{-13})(x + \sqrt{-13}) = (y)^3.$$ Since $P = (1)$, the ideals on the left hand side are coprime. Thus any primes dividing $(y)$ must divide either $(x + \sqrt{-13})$ or $(x - \sqrt{-13})$, but not both. Let $(y) = P_1,\ldots, P_mQ_1,\ldots, Q_n$, where the $P_i\mid(x + \sqrt{-13})$ and the $Q_i\mid (x - \sqrt{-13})$. Then from ($**$) it follows that $P_1^3\cdots P_m^3 = (x + \sqrt{-13})$ and $Q_1^3\cdots Q_n^3 = (x - \sqrt{-13})$. In particular, both $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$ are cubes of ideals.
Thanks so very much for such a good answer! There is just one question I have and that is right at the start when you proved P divides $(y)^3$. Why does this mean P divides $(y)$ as we are not assuming $P$ is a prime ideal? –  Alex Kite May 13 '12 at 0:01
@AlexKite: Actually, I think that was a mistake in my original answer. I probably shouldn't have used the letter $P$, since you start thinking of it as a prime ideal. We know that $P\mid(y)^3$, but that does not mean it divides $(y)$. The conclusion is the same however -- and I edited my answer to reflect these changes. Thanks for pointing it out! –  froggie May 13 '12 at 13:14