Show that $x^2 + x + 12 = 3y^5$ has no integer solutions.

Show that $x^2 + x + 12 = 3y^5$ has no integer solutions.

Use the fact that the class group of $K$ is cyclic of order 5, where $K=\mathbb{Q}[\alpha]$ and $\alpha$ is the root of $x^2-x+12$.

We get that $$\alpha = \frac{1+\sqrt{-47}}{2}$$

One can factor the LHS:

$$(x+\alpha)(x+\bar{\alpha})$$ and then what? If we show the above factors are coprime, saying that they are the 5th powers of some ideal doesn't help much, since we know that the 5th power of any ideal is a principal ideal due to the class group being $C_5$. -How does "3" come into play?

Solution 1:

Assuming you meant $K=\mathbb{Q}(\alpha)=\mathbb{Q}[x]/(x^2+x+12)$, then the first step is to factor as you did $$ (x+\alpha)(x+\overline{\alpha})=3y^5. $$ In fact $\overline{\alpha}=-(1+\alpha)$. It's helpful to consider the ideals generated by these elements, say $$ \begin{align*} I&=(x+\alpha), & I'&=(x+\overline{\alpha}), & J&=(y). \end{align*} $$ Coprime. Let $G=I+I'$ be the greatest common divisor. We want to show that $I$ and $I'$ are coprime, i.e. that $G=(1)$. Note that $1+2\alpha\in G$, and so $G\mid(1+2\alpha)$. Now $\operatorname{N}(1+2\alpha)=47$, which is a prime! In fact $\Delta_K=-47$, and so is the unique ramified prime of $K$: $(47)=\mathfrak{p}^2$, where $\mathfrak{p}=(1+2\alpha)$ is principal. Hence $G=\mathfrak{p}$ or $(1)$.

In the former case, $\mathfrak{p}\mid I,I'$, but no greater power of $\mathfrak{p}$ can (ideals have the same norm and $\mathfrak{p}$ is the unique prime above 47). This yields a square factor of $\mathfrak{p}$ in II', but a $5$-th power in $J^5$, which is a contradiction.

Using the class group. From the above, we can write $A^5\mid I$, $(A')^5\mid I'$, for some ideals $A$ and $A'$. $I$ and $I'$ aren't necessarily $5$-th powers themselves for the factor of $3$ that is involved. Since $(3)=(3,\alpha)(3,1+\alpha)$, those factors will need to appear somewhere.

As $h_K=5$, we know that $A^5$ and $(A')^5$ are principal, and as both $I$ and $I'$ are principal, either $I=(3)A^5$ and $I'=(A')^5$, or $I=A^5$ and $I'=(3)(A')^5$, as the factors of $(3)$ are not principal.

Suppose the former, and let $I'=(a+b\alpha)^5$. Then $$ \begin{align*} ((x-1)-\alpha)&=(a+b\alpha)^5, \\ &=(a^5+5a^4b\alpha+10a^3b^2\alpha^2+10a^2b^3\alpha^3+5ab^4\alpha^4+b^5\alpha^5), \\ &=(a^5+5a^4b\alpha-10a^3b^2(12+\alpha)+10a^2b^3(12-11\alpha)+5ab^4(132+23\alpha) \\ &\,\,\,\,\,\,\,\,-b^5(276-109\alpha)), \\ &=((a^5-120a^3b^2+120a^2b^3+660a^4b-276b^5) \\ &\,\,\,\,\,\,\,\,+(5a^4b-10a^3b^2-110a^2b^3+115ab^4+109b^5)\alpha). \end{align*} $$ Since $\mathcal{O}_K^\times=\langle -1\rangle$, equating coefficients from the generators of the above ideals we have $$ -1=\pm(5a^4b-10a^3b^2-110a^2b^3+115ab^4+109b^5), $$ or $$ -\frac{1}{b}=\pm(5a^4-10a^3b-110a^2b^2+115ab^3+109b^4), $$ which instantly forces $b=\pm 1$ as $a,b\in\mathbb{Z}$. At this point you need only convince yourself that this will have no solutions for $a\in\mathbb{Z}$, as then we're done (the later case where $(3)\mid I'$ is done similarly).

If you want to show this, you could do something along the lines of this. Suppose that $b=1$. Then $$ \pm 1=5a^4-10a^3-110a^2+115a+109=5(a-1)a((a-1)a-23)+109, $$ i.e. $$ -110=5(a-1)a((a-1)a-23)\text{ or }-108=5(a-1)a((a-1)a-23). $$ The second case cannot happen since $5\nmid 108$. For the first case, we require $a,a-1\mid 110$. The only pairs of consecutive integers dividing $110$ are $(1,2)$, $(10,11)$ and their negatives, so $a=-10,-1,2$ or $11$, all of which can quickly be ruled out by plugging in above. Similarly for $b=-1$.

Solution 2:

Assume $x,y\in \Bbb Z$ is a solution. Note that the gcd of $(x+\alpha)$ and $(x+\bar{\alpha})$ divides $(\alpha-\bar\alpha)=(\sqrt{-47})$. So the gcd is necessarily a principal ideal.

Let $ \mathfrak p \neq (\sqrt{-47})$ be a prime dividing $(y)$ in $\mathcal{O}_K=\Bbb Z[\alpha]$. Then $\mathfrak p$ divides exactly one the factors $(x+\alpha)$ and $(x+\bar \alpha)$. Thus $\mathfrak p^5$ divides this factor. Furthermore, $\mathfrak{p}^5$ is a principal ideal as $h_K = 5$.

So all factors of $y^5$ can be rearranged on the RHS to principal ideals dividing $(x+\alpha)$ or $(x+\bar \alpha)$.

Finally, the prime $(3)$ splits in $\mathcal O_K$ into $$(3) = (3,\alpha)(3,\bar \alpha) $$ and neither factor is a principal ideal. It is clear that $3 \not \mid x+\alpha,x+\bar \alpha $ in $\Bbb Z[\alpha]$ because the coefficient of $\alpha$ is not divisible by $3$. Therefore, each factor of $(3)$ divides either $(x+\alpha)$ or $(x+\bar \alpha)$.

Now we fit the pieces together: By the above consideration $(x+\alpha)$ factors into principal ideals and exactly one of $(3,\alpha)(3,\bar \alpha)$. Consequently, it can't be a principal ideal - a contradiction.