Diophantine equation $x^2 + 6(y+1)^2 = (y+2)^3$

Question

I'm revising for exams and I've got stuck on an algebraic number theory question. The equation I'm trying to solve is $$ x^2 -2 = y^3, $$ and I was told to rewrite it as $$ x^2 + 6(y+1)^2 = (y+2)^3. $$ Then I was told to show there exist integers $u,v$ such that $$ u^2 + 6v^2 = y+2 \text{ and } x +(y+1)\sqrt{-6} = (u+v\sqrt{-6})^3. $$ I tried factoring the equation in ideals as $$ \big( x + \sqrt{-6}(y+1)\big)\big( x - \sqrt{-6}(y+1)\big) = \big(y+2 \big)^3, $$ and I've been told for $K= \mathbb{Q}(\sqrt{-6})$ the only units are $\pm 1$ and the class number is $2$. But I can't show the two ideals on the left are coprime, so I'd really appreciate if someone could explain this to me. Thanks!

Answer 1

I think I figured this out for myself anyway, so I thought I'd write it up here.

So we're trying to show first that the ideals $$ \left( x + \sqrt{-6}(y+1) \right) \text{ and } \left( x - \sqrt{-6}(y+1) \right) $$ are coprime. So suppose there is some prime ideal $P$ dividing both of the ideals, then $2\sqrt{-6}(y+1) \in P$. Then we get $$ \left( 2\sqrt{-6}(y+1) \right) \subseteq P \text{, and so } P \mid \left(2\sqrt{-6}(y+1)\right). $$ Since $P$ is prime we can then deduce that either $$ P \mid (2), \text{ } P \mid\left(\sqrt{-6} \right) \text{ or } P \mid (y+1). $$ Suppose the last case is true, then since $$ \left( x + \sqrt{-6}(y+1) \right)\left( x - \sqrt{-6}(y+1) \right) = (y+2)^3, $$ $P$ also divides $(y+2)^3$, and so divides $(y+2)$. Then we have that $(y+2)-(y+1) = 1 \in P$, which contradicts that $P$ is a prime ideal.

Next suppose that $P$ divides $(\sqrt{-6})$ or $(2)$. Using Dedekind's theorem (I'm not going to write out all the details seeing as this was a question I asked), if we define $P_2 = (2, \sqrt{-6})$, $P_3 = (3, \sqrt{-6})$ then we have $$ (2) = P_2^2, \text{ and } (\sqrt{-6}) = P_2P_3. $$ First suppose $P_2$ divides both ideals, then $P_2 \mid( y+2)^3$, so $P_2 \mid (y+2)$, as $P_2$ is prime (again by Dedekind's theorem). Then by taking norms we get $$ 2 \mid y^2 + 4y + 4, \text{ in the integers}. $$ Thus $y = 2y'$ and similarly from the initial equation $x^2 - 2 = y^3$ we get $x = 2x'$ for some $x',y'$ in the integers. Then we get the equation $$ 2x'^2 - 1 = 4y'^2. $$ This equation clearly has no integer solutions upon reducing modulo $2$, so we get a contradiction and $P_2$ cannot divide both the ideals on the left hand side of the equation.

So lastly we need to consider $P_3$. Similarly we get, by taking norms, that $$ 3 \mid (y+2)^2, $$ and this gives ys that $y$ is congruent to $1 \pmod{3}$. Then considering the equation $x^2 - 2 = y^3$ gives that $x$ is congruent to $0$ modulo $3$. Substituting in $x = 3x'$ and $y=3y' + 1$ gives, upon expanding and dividing by $3$, $$ 3x'^2 - 1 = 9y'^3 + 9y'^2 + 3y', $$ which again has no solutions after reducing modulo 3. So $p_3$ cannot divide both the ideals either and so $\left( x + \sqrt{-6}(y+1) \right)$ and $\left( x - \sqrt{-6}(y+1) \right)$ are coprime ideals.

Since we have unique factorisation for ideals in a ring of integers we can then write $$ \left( x + \sqrt{-6}(y+1) \right) = I_1^3 \text{ and } \left( x - \sqrt{-6}(y+1) \right) = I_2^3 $$ for some ideals $I_1$ and $I_2$ of $\mathbb{Z}[\sqrt{-6}] = \mathcal{O}_K$ where $K = \mathbb{Q}(\sqrt{-6})$. Since the class number is $2$ and $I_2$ cubes to a principal ideal we must have that $I_1$ is principal, thus $$ \left( x + \sqrt{-6}(y+1) \right) = (\alpha)^3 = (\alpha^3), $$ for some $\alpha$ in $\mathbb{Z}[\sqrt{-6}]$. Since the only units in $\mathbb{Z}[\sqrt{-6}]$ are $\pm1$ we then get $$ x + \sqrt{-6}(y+1) = \pm \alpha^3 = (u +\sqrt{-6}v)^3, $$ for some $u,v$ in $\mathbb{Z}$. Then we can immediately see that $$ x - \sqrt{-6}(y+1) = (u - \sqrt{-6}v)^3, $$ and so $(y+2)^3 = (u^2 + 6v^2)^3$. This then gives us $$ y+2 = u^2 + 6v^2, $$ which was the end of the question I asked, but I'm going to write out the rest of the solution to the diophantine equation anyway, for completeness.

We can then expand $x + \sqrt{-6}(y+1) = (u + \sqrt{-6}v)^3$ and equate the coefficients of $1$ and $\sqrt{-6}$ to get $$ x = u^3 -18uv^2, \text{ and } y +1 = 3u^2v - 6v^3. $$ Then as $y+ 2 = u^2 + 6v^2$ we have $$ u^2 + 6v^2 = 1 + 3u^2v -6v^3, $$ which we can rearrange to $$ u^2 = \frac{ 1-6v^2(1+ v)}{1-3v}, \text{ or } 9u^2 = \frac{45v^2(1+v) - 9}{3v-1}. $$ Just by polynomial division this gives $$ 9u^2 = 18v^2 + 18v + 8 - \frac{1}{3v-1}, $$ and so $9u^2$ is an integer only if $\frac{1}{3v-1}$ is an integer which is true only if $v = 0$. Thus in particular $u$ is an integer only if $v = 0$. Then $v = 0$ gives us that $u^2 = 1 $ by substituting into the above equation. Thus, $y = -1$ and so $x^2 = 1$ and the only integer solutions to the equation are $$ (x,y) = (\pm1, -1). $$