Find all integer solutions for Mordell's Equation $x^2=y^3+k$, where $k=-35$.

Question

I know that there are definitely solutions to this equation, two of which are $(\pm 36,11)$. However, I now need some guidance on where to begin finding the remaining solutions for this case of Mordell's Equation.

Any help would be appreciated, thanks!

EDIT: These two are the only solutions, but I would still like to know a more rigorous way of finding them.

Answer 1

Note: I initially very foolishly misread the question, and typed up a solution for $k = 35$, instead of $-35$, only realising my mistake when I got to the end. Fortunately the argument is similar so I was able to salvage most of it. However, it is possible that I haven't quite changed all the $-$ signs to $+$ signs and the like, so please forgive any such errors.

Let $K = \mathbb{Q}(\sqrt{-35})$. The class group $\text{Cl}(K)$ of $K$ is isomorphic to $C_2$. This fact can be found here: https://www.lmfdb.org/NumberField/2.0.35.1, and for a proof see Class Group of $\mathbb Q(\sqrt{-35})$. If you're not familiar with class groups, they aren't much of a reach beyond unique prime factorisation of ideals, so it shouldn't take a lot of effort to gain sufficient understanding to follow the proof.

Note that the ring of integers of $K$ is $\mathcal{O}_K = \mathbb{Z}[\frac{1 +\sqrt{-35}}{2}]$; this follows from very standard results about quadratic number fields.

We are interested in integer pairs $(x, y)$ with $x^2+35 = y^3$. Suppose that such a pair exists. We will show that it must be one of the two you mention. Working over $\mathcal{O}_K$, we have $(x + \sqrt{-35})(x - \sqrt{-35}) = y^3$. The key idea here is pass into the following equation of ideals: $$ (x + \sqrt{-35})(x - \sqrt{-35}) = (y)^3 $$ The bracket notation for principal ideals is a bit confusing in this case, so I'll try to be clear verbally about when we are working with ideals, and when we are working with elements.

Now, we claim that the ideals $(x + \sqrt{-35})$ and $(x - \sqrt{-35})$ are actually coprime ideals. To see this, suppose that $\mathfrak{p}$ is an ideal dividing both of them. Then, as division implies containment, we have $x \pm \sqrt{-35} \in \mathfrak{p}$, and hence $2\sqrt{-35} \in \mathfrak{p}$, so $\mathfrak{p} \mid (2\sqrt{-35})$. Since norm is multiplicative, we have $N(\mathfrak{p}) \mid N((2\sqrt{-35})) = 2^2\cdot 5 \cdot 7$.

Note that also $\mathfrak{p} \mid (x + \sqrt{-35})$ and $(x + \sqrt{-35})\mid (y)^3$, so $\mathfrak{p} \mid (y)^3$, and hence $N(\mathfrak{p}) \mid N((y)^3) = y^6$. Since $\mathfrak{p}$ is a proper ideal of $\mathcal{O}_K$, its norm is a positive integer, so since it divides $2^2 \cdot 5 \cdot 7$, we have that $N(\mathfrak{p})$ must be divisible by one of the primes $2, 5, 7$ and hence $y^6$ is too. So $y$ is divisible by one of the primes $2, 5, 7$.

If $2 \mid y$, then $x^2 \equiv 3$ (mod $8$), which is impossible. If $p \mid y$ for $p = 5$ or $7$, then we get $x^2 \equiv -35$ (mod $p^3$) which is a contradiction because $p \mid x$ so $x^2$ should be a multiple of $p^2$.

Thus, $N(\mathfrak{p})$ cannot be divisible by any of $2, 5, 7$, which is a contradiction. Therefore, we conclude that the ideals $(x \pm \sqrt{-35})$ are indeed coprime. The product of these two coprime ideals is a cube, so it follows from unique prime factorisation of ideals (since their prime factorisations are disjoint) that each of $(x \pm \sqrt{-35})$ is a cube.

In particular, $(x + \sqrt{-35}) = \mathfrak{a}^3$ for some ideal $\mathfrak{a}$ of $\mathcal{O}_K$. This is where we invoke our earlier remarks about the class group. Since $(x + \sqrt{-35})$ is principal, we have that $[\mathfrak{a}]^3$ is trivial in $\mathcal{O}_K$. Since this group has order $2$, it follows that $[\mathfrak{a}]$ is trivial in $\text{Cl}(K)$, and therefore $\mathfrak{a}$ is a principal ideal, so we have $\mathfrak{a} = (a + b\frac{1 + \sqrt{-35}}{2})$ for some integers $a, b$. Expanding the generator of $\mathfrak{a}$ in the equation $\mathfrak{a} = (a + b\frac{1 + \sqrt{-35}}{2})$, we have

$$ (x + \sqrt{-35}) = (((a + \frac{b}{2})^3 - 3\cdot 35 (a + \frac{b}{2})(\frac{b}{2})^2 + (3(a + \frac{b}{2})^2(\frac{b}{2}) - 35(\frac{b}{2})^3))\sqrt{-35}) $$

And since the units of $\mathcal{O}_K$ are precisely $\{\pm 1\}$, we have (as elements, not ideals) $$ x + \sqrt{-35} = \pm (((a + \frac{b}{2})^3 - 3\cdot 35 (a + \frac{b}{2})(\frac{b}{2})^2 + (3(a + \frac{b}{2})^2(\frac{b}{2}) - 35(\frac{b}{2})^3))\sqrt{-35}) $$

So comparing coefficients of $\sqrt{-35}$, we have $$ 3(a + \frac{b}{2})^2(\frac{b}{2}) - 35(\frac{b}{2})^2 = \pm 1 $$ clearing denominators and rearranging, this becomes $$ b(3(2a + b)^2 - 35) = \pm 8 $$

This gives us a small number of cases to check (since $b$ must be a factor of $8$), and in fact we can speed this up by spotting that no matter the value of $b$, we need $3(2a + b)^2$ to be close to $35$ (specifically within a distance of $8$). The only number that is three times a square, and is less than $8$ away from $35$, is $27 = 3 \cdot 3^2$. This is $8$ away from $35$, so we need $b = \pm 1$. From here the casework really is trivial (and I don't want to do it), so I'll leave it at that.

Answer 2

In Sage:

R.<x,y> = QQ[]
E=EllipticCurve(y^3-x^2-35)
E.integral_points(both_signs=true)

$$[(11 : -36 : 1), (11 : 36 : 1)]$$