The exact formula conjecture for primitive representations by x^2+216y^2

Write

$$m=\prod_{i=1}^k p_i^{a_i} \prod_{j=1}^n q_j^{b_j} \equiv1\pmod{24},$$

with $p_i\in\mathcal P$, $q_j\notin\mathcal P$, and every prime divisor congruent to $1,5,7,$ or $11\pmod{24}$. Put

$$B=\#\{j:3\mid b_j\}, \qquad t=n-B.$$

We prove the exact conjectured formula

$$\boxed{ N_2(m) = \frac{2^{n+k+1} + 2^{B+k+2}(-1)^{n-B}}3. }$$

Consider the quadratic order

$$\mathcal O=\mathbb Z[\sqrt{-216}] =\mathbb Z[6\sqrt{-6}], \qquad \operatorname{disc}(\mathcal O)=-864.$$

Its twelve reduced primitive positive forms fall into the following four genera, labeled by the residue modulo $24$ that they represent coprimely to $6$:

$$\begin{array}{c|l} 1 &[1,0,216],\ [9,-6,25],\ [9,6,25]\\ 5 &[5,-4,44],\ [5,4,44],\ [8,8,29]\\ 7 &[4,4,55],\ [7,-2,31],\ [7,2,31]\\ 11&[8,0,27],\ [11,-4,20],\ [11,4,20]. \end{array}$$

The four ambiguous classes are

$$[1,0,216],\quad[4,4,55], \quad[8,0,27],\quad[8,8,29].$$

Consequently,

$$\operatorname{Cl}(\mathcal O) \cong C_2\times C_2\times C_3.$$

Every prime dividing $m$ is coprime to the conductor and satisfies

$$\left(\frac{-6}{p}\right)=1.$$

It therefore splits into conjugate invertible prime ideals. Choose one such ideal $\mathfrak p$, and write

$$[\mathfrak p] =(\gamma_p,\epsilon_p) \in(C_2\times C_2)\times C_3.$$

The characterization of $\mathcal P$ by primitive representations of $p^2$ gives

$$p\in\mathcal P \iff[\mathfrak p]^2=1 \iff\epsilon_p=0.$$

A primitive representation

$$m=x^2+216y^2, \qquad\gcd(x,y)=1,$$

selects, for each prime power $p^e\Vert m$, exactly one of the two conjugate primes, to its full exponent. Thus candidate primitive ideals correspond to independent signs

$$\delta_p\in\{1,-1\},$$

and their classes have coordinates

$$\left( \prod_{p^e\Vert m}\gamma_p^e,\ \sum_{p^e\Vert m}\delta_pe\epsilon_p \right).$$

Because $m\equiv1\pmod{24}$, the genus coordinate is trivial for every choice of signs. Therefore the corresponding ideal is principal precisely when

$$\sum_{p^e\Vert m}\delta_pe\epsilon_p=0 \quad\text{in }C_3.$$

The $k$ primes in $\mathcal P$ each contribute an unrestricted factor $2$. So do the $B$ primes outside $\mathcal P$ whose exponents are divisible by $3$. The remaining $t=n-B$ primes contribute signs in $C_3$, and the root-of-unity filter gives

$$\#\left\{ (\delta_1,\ldots,\delta_t)\in\{\pm1\}^t: \sum_j\delta_j=0\pmod3 \right\} = \frac{2^t+2(-1)^t}{3}.$$

Finally,

$$\mathcal O^\times=\{\pm1\},$$

so each principal ideal gives exactly two primitive representations. Hence

$$N_2(m) = 2^{k+B+1} \frac{2^{n-B}+2(-1)^{n-B}}3 = \frac{2^{n+k+1} + 2^{B+k+2}(-1)^{n-B}}3,$$

as claimed. The empty-factor case $m=1$ is included and gives $N_2(1)=2$.