A modulo 192 congruence family for partitions with distinct even parts

A harmonic-theta proof of the distinct-even-parts congruence modulo 192

Problem. MathDB #367436.

Primary source. H. Nath and A. Sarma, Congruences and density results for partitions into distinct even parts, Acta Mathematica Hungarica 177 (2025), 140–162, doi:10.1007/s10474-025-01573-9; arXiv:2503.06228, Conjecture 7.1, equation (1.2), and Lemma 5.2. The source already proves its different Conjecture 1.1 in Theorem 1.2; the result below concerns only the subsequent, previously unresolved Conjecture 7.1.

Theorem

Let $p\geq5$ be prime and suppose that

$$\left(\frac{-2}{p}\right)=-1. \tag{1}$$

Let $t\geq1$ satisfy $\gcd(t,6)=1$ and $p\mid t$. Then, for every $n\geq0$ and every $1\leq j\leq p-1$,

$$\boxed{ \operatorname{ped}\left( 9t^2n+\frac{9t^2j}{p}+\frac{57t^2-1}{8} \right) \equiv0\pmod{192}. } \tag{2}$$

In fact, the following stronger coefficient-support statement holds:

$$\boxed{ \operatorname{ped}(9k+7)\equiv0\pmod{192} \quad\text{whenever }24k+19 \text{ has an inert prime divisor to odd valuation.} } \tag{3}$$

Here a prime is inert precisely when it is congruent to $5$ or $7$ modulo $8$.

1. The source generating function and its modular lift

Write

$$f_r(q)=\prod_{m\geq1}(1-q^{rm}).$$

Equation (1.2) of the primary source gives the exact integral identity

$$\sum_{k\geq0}\operatorname{ped}(9k+7)q^k =12A(q), \qquad A(q)=\frac{f_2(q)^4f_3(q)^6f_4(q)}{f_1(q)^{11}}. \tag{4}$$

Introduce

$$B(q)=\frac{f_1(q)^5f_3(q)^6f_4(q)}{f_2(q)^4}. \tag{5}$$

The ratio of these two integral power series is

$$\frac{B(q)}{A(q)} =\frac{f_1(q)^{16}}{f_2(q)^8} =\prod_{m\geq1} \left(\frac{1-q^m}{1+q^m}\right)^8. \tag{6}$$

For an indeterminate $X$,

$$(1-X)^8-(1+X)^8 =-16\left(X+7X^3+7X^5+X^7\right). \tag{7}$$

Since $(1+X)^8$ is a unit in $\mathbb Z[[X]]$, equations (6) and (7) imply

$$A(q)\equiv B(q)\pmod{16}. \tag{8}$$

Thus, with $q=e^{2\pi iz}$, define the integral modular form

$$E(z) =q^{19}B(q^{24}) =\frac{\eta(24z)^5\eta(72z)^6\eta(96z)}{\eta(48z)^4}. \tag{9}$$

The source's Lemma 5.2 with its parameter $k=3$ states that this eta quotient is holomorphic of weight $4$ on $\Gamma_0(2304)$. Its character is trivial: the product of eta arguments with their multiplicities is a square, since its $2$-adic and $3$-adic valuations are respectively $22$ and $14$. In particular,

$$E\in M_4\bigl(\Gamma_1(4608)\bigr). \tag{10}$$

Combining the exact identity (4) with the independently established modulo-$16$ congruence (8), rather than dividing a congruence by $12$, gives

$$\operatorname{ped}(9k+7) \equiv12[q^{24k+19}]E(z)\pmod{192}. \tag{11}$$

2. An explicit harmonic theta series

For odd integers $u$, put

$$\chi_{-8}(u)=\left(\frac{-2}{u}\right), \qquad \chi_8(u)=\left(\frac{2}{u}\right). \tag{12}$$

The two characters have respective values

$$\begin{array}{c|rrrr} u\bmod8&1&3&5&7\\ \hline \chi_{-8}(u)&1&1&-1&-1\\ \chi_8(u)&1&-1&-1&1. \end{array} \tag{13}$$

Consider the positive-definite quadratic form and cubic polynomial

$$Q(x,y)=x^2+2y^2, \qquad H(x,y)=x^3-6xy^2. \tag{14}$$

The polynomial is harmonic for this quadratic form:

$$\left( \frac{\partial^2}{\partial x^2} +\frac12\frac{\partial^2}{\partial y^2} \right)H(x,y) =6x-6x=0. \tag{15}$$

Define

$$\Theta(z) =\frac14 \sum_{\substack{(x,y)\in\mathbb Z^2\\ \gcd(x,6)=1\\ y\equiv3\, (\mathrm{mod}\,6)}} \chi_{-8}(x)H(x,y) \left(11-4\chi_8(x)+4\chi_8(y)\right) q^{Q(x,y)}. \tag{16}$$

The summand is invariant under each independent sign change $x\mapsto-x$ and $y\mapsto-y$: the odd character $\chi_{-8}(x)$ and the odd polynomial $H(x,y)$ change sign together, while $\chi_8$ is even. Neither coordinate can vanish under the displayed restrictions. Therefore every sign orbit has exactly four elements, which cancel the prefactor $1/4$ integrally. Equivalently,

$$\Theta(z) =\sum_{\substack{x,y\geq1\\ \gcd(x,6)=1\\ y\equiv3\, (\mathrm{mod}\,6)}} \chi_{-8}(x)H(x,y) \left(11-4\chi_8(x)+4\chi_8(y)\right) q^{x^2+2y^2} \in\mathbb Z[[q]]. \tag{17}$$

Moreover, every exponent in (17) satisfies

$$x^2+2y^2\equiv3\pmod8, \qquad x^2+2y^2\equiv1\pmod3,$$

and consequently

$$x^2+2y^2\equiv19\pmod{24}. \tag{18}$$

Conversely, if $x^2+2y^2\equiv19\pmod{24}$, then $x$ and $y$ are odd, $3\nmid x$, and $3\mid y$. Hence (17) includes every positive representation of each exponent $24k+19$.

3. Modularity on the correct congruence subgroup

The weight in (16) is periodic modulo $24$ in both coordinates. Take the even lattice

$$L=24\mathbb Z^2, \qquad (u,v)=2u_1v_1+4u_2v_2. \tag{19}$$

In its standard lattice basis, the Gram matrix is

$$G= \begin{pmatrix} 1152&0\\ 0&2304 \end{pmatrix}. \tag{20}$$

Its level is the least positive $N$ for which $NG^{-1}$ is integral with even diagonal. Thus

$$N=4608. \tag{21}$$

Every residue class $h=(r,s)+L$ used in (16) belongs to $L^*/L$, since

$$\bigl((r,s),(24a,24b)\bigr) =48ra+96sb\in\mathbb Z. \tag{22}$$

The standard harmonic-theta transformation theorem makes

$$\theta_h(z) =\sum_{v\in h+L}H(v)q^{Q(v)} \tag{23}$$

a component of a vector-valued modular form of weight

$$\frac{\operatorname{rank}L}{2}+\deg H =\frac22+3=4 \tag{24}$$

for the Weil representation of $L^*/L$. Since $L$ is positive definite, all these components are holomorphic at the cusps.

The distinction between $\Gamma_1(N)$ and $\Gamma(N)$ is essential here. Because the lattice has even signature, its Weil representation factors through

$$\operatorname{SL}_2(\mathbb Z)/\Gamma(N) \cong\operatorname{SL}_2(\mathbb Z/N\mathbb Z). \tag{25}$$

This even-signature factorization is stated explicitly in M. K.-H. Müller and N. R. Scheithauer, The invariants of the Weil representation of $\operatorname{SL}_2(\mathbb Z)$. If $\gamma\in\Gamma_1(N)$, its reduction modulo $N$ is $T^b$ for some $b$, where

$$T= \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}. \tag{26}$$

The action on the basis vector indexed by $h$ is

$$\rho_L(T)e_h=e^{2\pi iQ(h)}e_h. \tag{27}$$

For the classes occurring in (16), $h$ has an integral representative $(r,s)$, so

$$Q(h)=r^2+2s^2\equiv0\pmod1. \tag{28}$$

Hence $\rho_L(T)^be_h=e_h$. The selected components (23) are therefore individually modular on $\Gamma_1(4608)$, even though arbitrary shifted components need not have this property. As (16) is a rational linear combination of those components,

$$\Theta\in M_4\bigl(\Gamma_1(4608)\bigr). \tag{29}$$

4. A complete exact Sturm certificate

The index of the relevant congruence subgroup is

$$\begin{aligned} \bigl[\operatorname{SL}_2(\mathbb Z):\Gamma_1(4608)\bigr] &=4608^2 \left(1-\frac1{2^2}\right) \left(1-\frac1{3^2}\right)\\ &=14{,}155{,}776. \end{aligned} \tag{30}$$

For weight $4$, Sturm's congruence bound is therefore

$$\mathcal B =\frac4{12}\cdot14{,}155{,}776 =4{,}718{,}592. \tag{31}$$

Both $E$ and $\Theta$ have integral coefficients and are supported on exponents congruent to $19$ modulo $24$. Consequently it suffices to check

$$[q^{24k+19}]E \equiv[q^{24k+19}]\Theta\pmod{16} \qquad \left( 0\leq k\leq \left\lfloor\frac{\mathcal B-19}{24}\right\rfloor =196{,}607 \right). \tag{32}$$

This is a finite, exhaustive certificate, not an asymptotic extrapolation. The accompanying dependency-free C++ verifier calculates the source partition numbers by the exact recurrence obtained from

$$f_1(q)\sum_{m\geq0}\operatorname{ped}(m)q^m=f_4(q), \tag{33}$$

using Euler's pentagonal theorem modulo $192$. Independently, it enumerates every positive representation

$$x^2+2y^2=24k+19$$

and evaluates the harmonic weight in (17) modulo $16$. It verifies (32) including its final endpoint, with integer arithmetic only.

Sturm's theorem applied modulo the ideal $(16)$ now gives the full formal power-series identity

$$\boxed{ E(z)\equiv\Theta(z)\pmod{16}. } \tag{34}$$

In particular, equations (11), (17), and (34) prove the stronger explicit formula

$$\boxed{ \begin{aligned} \frac{\operatorname{ped}(9k+7)}{12} &\equiv \sum_{\substack{x,y\geq1\\x^2+2y^2=24k+19}} \chi_{-8}(x)(x^3-6xy^2)\\ &\hspace{12mm}\cdot \left(11-4\chi_8(x)+4\chi_8(y)\right) \pmod{16}. \end{aligned} } \tag{35}$$

The quotient on the left is an integer by the exact identity (4).

5. The inert-prime obstruction

Suppose $p$ satisfies (1) and

$$p\mid x^2+2y^2.$$

If $p\nmid y$, reduction modulo $p$ would give

$$\left(xy^{-1}\right)^2\equiv-2\pmod p,$$

contradicting (1). Hence $p\mid y$, and then also $p\mid x$. Therefore

$$p^2\mid x^2+2y^2.$$

Iteration proves that

$$v_p(x^2+2y^2)\text{ is even for every }(x,y)\neq(0,0). \tag{36}$$

Thus the representation sum in (35) is empty whenever $24k+19$ has an inert prime to odd valuation. Equations (11) and (34) then give (3).

Finally, let the argument in the original conjecture be

$$M=9t^2n+\frac{9t^2j}{p}+\frac{57t^2-1}{8}. \tag{37}$$

Because $\gcd(t,6)=1$, we have $t^2\equiv1\pmod{24}$. Since $p\mid t$, the integer $M$ can therefore be written as

$$M=9k+7, \qquad k=t^2n+\frac{t^2j}{p}+\frac{19(t^2-1)}{24}. \tag{38}$$

Its associated norm parameter satisfies

$$\begin{aligned} 24k+19 &=t^2(24n+19)+\frac{24t^2j}{p}\\ &=\frac{t^2}{p} \left(p(24n+19)+24j\right). \end{aligned} \tag{39}$$

Since $p\geq5$ and $1\leq j\leq p-1$,

$$p\nmid p(24n+19)+24j.$$

Consequently,

$$v_p(24k+19)=2v_p(t)-1, \tag{40}$$

which is odd even when $p^2\mid t$. Applying (3) to (38) proves (2) for all the parameters in the source conjecture.

Conclusion. Source Conjecture 7.1 is PROVED in its full stated scope. Equation (35) additionally gives a weighted binary-quadratic-form formula for the complete progression $\operatorname{ped}(9k+7)$ modulo $192$.