Determine the number of solutions of the Diophantine equation $x^2+xy+6y^2=k$, for each $k\in\{2^{10},3^{11},59^{12},61^{13}\}$ Attempt:
First observe that $x^2+xy+6y^2=(x+y/2)^2+23y^2/4$ which is a norm of $x+y\frac{1+\sqrt{-23}}{2}\in \mathcal O_K$ the ring of integers of $K=\mathbb Q(\sqrt{-23})$ with integral basis $\{1,(1+\sqrt{-23})/2\}$.
So question reduces to show how many elements in $\mathcal O_K$ with norm $\in\{2^{10},3^{11},59^{12},61^{13}\}$
I know the class number of $\mathcal O_K$ is $3$ and $Cl(K)=\{1,[\mathfrak p_2],[\mathfrak p_2]^2\}\cong \mathbb Z_3$
where $\mathfrak p_2=\left(2,\frac{1+\sqrt{-23}}{2}\right)$
However I cannot conclude.

We are asked if there are principal ideals $\mathfrak{a}, \overline{\mathfrak{a}}$, such that $\mathfrak{a}\overline{\mathfrak{a}}=61^{13}$.
By Dedekind-Kummer theorem we have:
$$\left(\frac{-23}{61}\right)=-1.$$
Even if $\mathfrak{a}=\overline{\mathfrak{a}}$, this is still impossible, since by unique factorization: $\mathfrak{a}^2=\mathfrak{p}^{13}$ don't even match the exponents.
59 splits : 59=$\mathfrak{p}_1\mathfrak{p}_2$ and $\mathfrak{p}_1=(5+2\frac{1+\sqrt{-23}}{2})$. This is the only solution since the unit group $U=\{1,-1\}$.
3 splits, but it's decomposition is non-principal, thus raising to the power 11 leaves you with non-principal ideals (class number is 3).
