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?
 A: 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.
