Lets find all the integral solutions to \(x^3=y^2+74\) assuming that \(h_K=10\) where \(K=\mathbb {Q}(\sqrt{-74})\). The trick here is to factor this expression in \(\mathcal{O}_K=\mathbb {Z}[\sqrt{-74}]\), but before that, note the following:

Assume \(x,y \in \mathbb {Z}\) are solutions. First thing to note is that \(y\) cannot be even, since this would mean \(x\) is even. But then \(x^3 -y^2 \equiv 0 \pmod4\) but \(4\nmid 74\). So \(x,y\) must be odd. Similarly, we see that \(x,y\) must be coprime, since if \(p\) divides both then \(p^2\) divided \(x^3-y^2=74\) but \(74\) is square-free.

Now lets look at the ideals. We have

as ideals. Now, are the two ideals on the right hand side coprime?

If \(\mathfrak {p}\) divided them both, we would have

and therefore \(\mathfrak {p}\) divides their sum, so \(\mathfrak {p}| (2y)\). Looking at the other side of the equation, we see that \(\mathfrak {p}|(x)\). We know \(x\) is odd, so \(\mathfrak {p}\) cannot divide \((2)\) (if it did, then we would have \((x)=\mathfrak {p}_2 \mathfrak {a}\) where \(\mathfrak {a}\) is some ideal and \(\mathfrak {p}_2^2=(2)\) (as can be seen from Theorem 3.5.11), but this means \(N(\mathfrak {p}_2)=2\) which would mean \(N((x))=|N_{K/\mathbb {Q}}(x)|=x^2\) is even, which can’t happen as \(x\) is odd). So we have \(\mathfrak {p}\mid (y)\) and \(\mathfrak {p}|(x)\) but we know \(x,y\) are coprime (as integers, which means their ideals are also coprime). Therefore we cannot have a \(\mathfrak {p}\) dividing both factors on the right, so they are coprime.

Now, using Lemma 5.0.2 we must have \((y-\sqrt{-74})=\mathfrak {a}^3\) and \((y+\sqrt{-74})=\mathfrak {b}^3\) with \(\mathfrak {a}\mathfrak {b}=(x)\). Note this means \([\mathfrak {a}^3]=[1]=[\mathfrak {b}^3]\). But note the class group has size \(10\) so we cant have ideal classes of order \(3\). Therefore \(\mathfrak {a},\mathfrak {b}\) are principal. So let \(\mathfrak {a}=(a+b\sqrt{-74})\).

Then we have

Using Lemma 5.0.1 and Exercise 3.1.5 we see that up to \(\pm 1\) we have

Equating each side we have

The second equation implies \(b \mid -1\) and therefore \(b=\pm 1\). This then gives \(a=\pm 5\). Substituting, it then follows that \(y=\pm 985\) and \(x=99\) are the only solutions.

Lets do another example. Lets show that \(x^3=y^2+5\) has no integral solutions. We saw in Example 4.1.3 that the class group in this case has size \(2\).

Now, as we did before, lets start by assuming \(x,y\) are integer solutions to this equation. If \(x\) was even then \(y^2+5 \equiv 0 \mod 4\) which would make \(-1\) a square modulo \(4\) which can’t happen, so \(x\) cannot be even. Similarly, \(x,y\) must be coprime since if \(p\) divided both of them, \(p^2| (x^3-y^2)\) meaning \(p^2 \mid 5\) which is a contradiction.

Ok, with this done, lets now factor this in \(\mathcal{O}_K\) as ideals where \(K=\mathbb {Q}(\sqrt{-5})\). We have

Next, lets see if the terms on the right are coprime. If a prime ideal \(\mathfrak {p}\) divided both of them, then \(\mathfrak {p}\) would divide their sum \((2y)\) so \(\mathfrak {p}\mid (2y)\). Looking at the left of the equation we see \(\mathfrak {p}\) divides \((x)\) and therefore as in the previous example, since \(x\) is odd, we cant have \(\mathfrak {p}\) dividing \((2)\). But then \(\mathfrak {p}\mid (y)\) and \(\mathfrak {p}\mid (x)\) but this also cant happen as \(x,y\) are coprime.

So, using Lemma 5.0.2 we see that we must have \(\mathfrak {a}^3=(y-\sqrt{-5})\) and \(\mathfrak {b}^3=(y+\sqrt{-5})\) for some ideals \(\mathfrak {a},\mathfrak {b}\). Now, if we think of what the class group having size \(2\) says, it means that in particular, we cannot have any non-principal ideals whose cube is principal, since this would give an element of order \(3\) in the class group. Therefore, both \(\mathfrak {a}\) and \(\mathfrak {b}\) must be themselves principal.

So lets just look as \(\mathfrak {a}=a+b\sqrt{-5}\). We must have

which means there is some unit \(u \in \mathcal{O}_K^\times \) such that

now as elements, ot just ideals. Since \(\mathcal{O}_K^\times =\{ \pm 1\} \) as we saw in sheet 6, wlog we can assume

(since \(-1\) is a cube). If we expand out we get

Now, equating each side we have \(y=a^3-15ab^2\) and \(-1=3a^2b-5b^3\). The second equation tells us \(b \mid -1\) so \(b=\pm 1\). If we now, substitute this back into the first equation we get \(-1=3a^2-5\) which has no solution for \(a \in \mathbb {Z}\). So \(x^3=y^2+5\) has no integer solutions.