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

About 4 years old · traced to 2022

Let $m$ be a positive integer with $m\equiv 1\pmod{24}$ . Suppose that all prime divisors of $m$ are congruent to $1$ , $5$ , $7$ , or $11$ modulo $24$ , and write

m=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}q_1^{b_1}q_2^{b_2}\cdots q_n^{b_n},

where $p_1,\ldots,p_k$ belong to $\mathcal P$ and $q_1,\ldots,q_n$ belong to the set $\mathcal Q$ of primes not in $\mathcal P$ . Let $N_2(m)$ be the number of primitive integer solutions of $x^2+216y^2=m$ , and let $B(m):=|\\{b_i:b_i\equiv 0\pmod 3\\}|$ . The exact formula conjecture. With this notation,

N_2(m)=2^{B(m)+k}\cdot\frac{2^{n-B(m)+1}+4\cdot(-1)^{n-B(m)}}{3} =\frac{2^{n+k+1}+2^{B(m)+k+2}\cdot(-1)^{n-B(m)}}{3}.

The formula would give an exact count of the primitive representations relevant to the paper's Diophantine analysis; the source notes that it implies the main theorem, but provides no proof of the conjecture.

References

Primary source

Cristina Ballantine, Mircea Merca and Cristian-Silviu Radu, “Parity of 3-regular partition numbers and Diophantine equations”, arXiv:2212.09810 (2022).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

The conjecture gives an exact count for primitive representations by $x^2+216y^2$ ; an unverified complete-proof attempt has appeared, but no independent confirmation is recorded.

Ballantine, Merca, and Radu proposed the formula in 2022 for integers $m\equiv 1\pmod{24}$ with the stated prime-factor restrictions. Their paper says the formula would sharpen the counting result used in their Diophantine analysis, but does not prove it.

Known results

Ballantine, Merca, and Radu (2022) proved a weaker representation-counting identity sufficient for their partition-congruence theorem, rather than the conjectured exact formula.
They identified the obstruction that the relevant discriminant $-864$ is not idoneal, preventing their proof from separating all representation classes explicitly.

Posted attempt

A complete proof attempt claims that the quadratic order of discriminant $-864$ has class group $C_2\times C_2\times C_3$ and that an ideal-class root-of-unity count yields the stated formula, including $m=1$ . The argument has not been independently verified, so it establishes only claimed progress.

Current status (as of August 2026): the exact formula remains unverified; the published source gives only a weaker result, while a reader-written complete-proof attempt has no independent confirmation.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

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 $N_2(1)=2$ .