# Parity of $3$-regular partition numbers and Diophantine equations

Cristina Ballantine  Department of Mathematics and Computer Science
College of The Holy Cross
Worcester, MA 01610, USA<EMAIL_ADDRESS>,  Mircea Merca  Department
of Mathematical Methods and Models
Fundamental Sciences Applied in Engineering Research Center
University Politehnica of Bucharest
RO-060042 Bucharest, Romania<EMAIL_ADDRESS>and  Cristian-
Silviu Radu  Research Institute for Symbolic Computation (RISC), Johannes
Kepler University
4040 Linz, Austria<EMAIL_ADDRESS>
###### Abstract.

Let $b_{3}(n)$ be the number of $3$-regular partitions of $n$. Recently, W. J.
Keith and F. Zanello discovered infinite families of Ramanujan type
congruences modulo $2$ for $b_{3}(2n)$ involving every prime $p$ with $p\equiv
13,17,19,23\pmod{24}$, and O. X. M. Yao provided new infinite families of
Ramanujan type congruences modulo $2$ for $b_{3}(2n)$ involving every prime
$p\geqslant 5$. In this paper, we introduce new infinite Ramanujan type
congruences modulo $2$ for $b_{3}(2n)$. They complement naturally the results
of Keith-Zanello and Yao and involve primes in $\mathcal{P}=\\{p\text{ prime
}:\exists\,j\in\\{1,4,8\\},\,x,y\in\mathbb{Z},\,\gcd(x,y)=1\text{ with
}x^{2}+216y^{2}=jp\\}$ whose Dirichlet density is $1/6$. As a key ingredient
in our proof we show that of the number of primitive solutions for
$x^{2}+216y^{2}=pm$, $p\in\mathcal{P}$, $p\nmid m$ and $pm\equiv 1\pmod{24}$,
is divisible by $8$. Here, the difficulty arises from the fact that $216$ is
not idoneal. We also give a conjectural exact formula for the number of
solutions for this Diophantine equation. In the second part of the article, we
study reversals of Euler-type identities. These are motivated by recent work
of the second author on a reversal of Schur’s identity which involves
$3$-regular partitions weighted by the parity of their length.

Keywords: Diophantine equations, density of prime subsets, modular forms,
partition congruences, regular partitions

MSC 2020: 11D09, 11D45, 11P83, 05A17, 11F33

Cristian-Silviu Radu was supported by grant SFB F50-06 of the FWF

##  1\. Introduction

A partition of a positive integer $n$ is a weakly decreasing sequence of
positive integers whose sum is $n$. The positive integers in the sequence are
called parts. For more on the theory of partitions, we refer the reader to
[2].

For an integer $\ell>1$ a partition is called $\ell$-regular if none of its
parts is divisible by $\ell$. The number of the $\ell$-regular partitions of
$n$ is usually denoted by $b_{\ell}(n)$ and its arithmetic properties were
investigated extensively. See, for example, [3, 6, 7, 9, 11, 14, 16, 17, 23,
24, 22]. In classical representation theory, when $\ell$ is prime,
$\ell$-regular partitions of $n$ parameterize the irreducible $\ell$-modular
representations of the symmetric group $\mathcal{S}_{n}$ [12].

Recently, W. J. Keith and F. Zanello [13] studied the parity of the
coefficients of certain eta-quotients and investigated the parity of
$b_{\ell}(n)$ when $\ell\leqslant 28$. In particular, they proved the
following parity result for $b_{3}(n)$ which involves every prime $p$
satisfying $p\equiv 13,17,19,23\pmod{24}$.

######  Theorem 1.1 (Keith-Zanello).

The sequence $b_{3}(2n)$ is lacunary modulo $2$. If $p\equiv
13,17,19,23\pmod{24}$ is prime, then

$b_{3}\big{(}2(p^{2}n+pk-24^{-1})\big{)}\equiv 0\pmod{2}$

for $1\leqslant k\leqslant p-1$, where $24^{-1}$ is taken modulo $p^{2}$.

Motivated by the Keith-Zanello result, O. X. M. Yao [25] proved new infinite
families of congruences modulo $2$ for $b_{3}(2n)$. These new congruences
involve any prime $p\geqslant 5$.

######  Theorem 1.2 (Yao).

Let $p\geqslant 5$ be a prime.

  1. (1)

If $b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv 1\pmod{2}$, then for
$n,k\geqslant 0$

  2. $b_{3}\left(2p^{4k+4}\,n+2p^{4k+3}\,j+\frac{p^{4k+4}-1}{12}\right)\equiv 0\pmod{2}$

where $1\leqslant j\leqslant p-1$ and for $n,k\geqslant 0$

  3. $b_{3}\left(\frac{p^{4k}-1}{12}\right)\equiv 1\pmod{2}.$

  4. (2)

If $b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv 0\pmod{2}$, then for
$n,k\geqslant 0$ with $p\nmid(24n+1)$

  5. $b_{3}\left(2p^{6k+2}\,n+\frac{p^{6k+2}-1}{12}\right)\equiv 0\pmod{2}$

and for $n,k\geqslant 0$

  6. $b_{3}\left(\frac{p^{6k}-1}{12}\right)\equiv 1\pmod{2}.$

In this paper, motivated by Theorems 1.1 and 1.2, we provide new Ramanujan
type congruences modulo $2$ for $b_{3}(2n)$. First we introduce some notation.
Given the Diophantine equation $x^{2}+ny^{2}=m$, by a primitive solution we
mean a solution $(x,y)\in\mathbb{Z}^{2}$ satisfying $\gcd(x,y)=1$.

For the remainder of the article $\mathcal{P}$ denotes the set of primes $p$
such that for some $j\in\\{1,4,8\\}$ the equation $x^{2}+24\cdot 9y^{2}=jp$
has primitive solutions.

######  Theorem 1.3.

For every $p\in\mathcal{P}$ and $n\geqslant 0$, we have

$b_{3}\big{(}2\,(p^{2}\,n+p\,\alpha-24_{p}^{-1})\big{)}\equiv 0\pmod{2},$

where $0\leqslant\alpha<p$, $\alpha\neq\lfloor p/24\rfloor$, $24_{p}^{-1}$ is
the inverse of $24$ modulo $p$ taken such that
$1\leqslant-24_{p}^{-1}\leqslant p-1$.

######  Remark 1.

In the proof of Theorem 1.1, the authors make use of the fact that if $p$ is a
prime congruent to $13,17,19,23$ modulo $24$, then the Diophantine equation
$x^{2}+24y^{2}=pm$ has no solutions. In section 2.2 we show that, if
$p\in\mathcal{P}$, then

$\displaystyle j=1$ $\displaystyle\implies p\equiv 1\pmod{24}$ $\displaystyle
j=4$ $\displaystyle\implies p\equiv 7\pmod{24}$ $\displaystyle j=8$
$\displaystyle\implies p\equiv 5,11\pmod{24}.$

For $p\in\mathcal{P}$, the equation $x^{2}+24y^{2}=pm$ may have solutions and
we need to understand the behavior modulo $8$ of the number of solutions with
$3\nmid y$. Here, $pm=24((p^{2}\,n+p\,\alpha-24_{p}^{-1}))+1$.

######  Remark 2.

It will become clear from the discussion in section 2.2 that for
$p\in\mathcal{P}$ we have $b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv
1\pmod{2}$.

Thus, Theorem 1.3 complements naturally Theorems 1.1 and 1.2.

A few words about the proof of Theorem 1.3 are in order. It follows from [13]
that $b_{3}(2n)\equiv a(n)\pmod{2}$, where $a(n)$ is the number of
representations of $n$ as the sum of a generalized pentagonal number and a
square not divisible by $3$. Thus, to prove Theorem 1.3 we have to investigate
the number of solutions $(x,y)\in\mathbb{N}^{2}$ for

(1) $\ x^{2}+24y^{2}=pm$

with $y=0$ or $3\nmid y$, and $pm$ as in Remark 1. This in turn, requires a
study of the number of primitive solutions for $x^{2}+216y^{2}=pm$ which we
denote by $N_{2}(pm)$.

Regarding (1), we note that there are four positive, reduced, primitive
quadratic forms of discriminant $-96$. We denote by $N(-96,w)$ the number of
primitive representations of $w$ by all these quadratic forms, i.e.,

$\begin{split}N(-96,w):=&|\\{(x,y)\in\mathbb{Z}:x^{2}+24y^{2}=w,\,\gcd(x,y)=1\\}|\\\
+&|\\{(x,y)\in\mathbb{Z}:3x^{2}+8y^{2}=w,\,\gcd(x,y)=1\\}|\\\
+&|\\{(x,y)\in\mathbb{Z}:5x^{2}+2xy+5y^{2}=w,\,\gcd(x,y)=1\\}|\\\
+&|\\{(x,y)\in\mathbb{Z}:4x^{2}+4xy+7y^{2}=w,\,\gcd(x,y)=1\\}|.\end{split}$

Then the mass formula (see, for example [5, Lemma 3.25]) gives

$N(-96,w)=2\prod_{p\mid w}\left(1+\Big{(}\frac{-6}{p}\Big{)}\right),$

where the product is over all prime divisors of $w$ and
$\big{(}\frac{-6}{p}\big{)}$ is the Jacobi symbol.

However, since $\alpha^{2}\equiv 1\pmod{24}$ when $\gcd(\alpha,6)=1$, we see
that, when $w\equiv 1\pmod{24}$, there are no integers $x,y$ such that
$3x^{2}+8y^{2}=w$, $5x^{2}+2xy+5y^{2}=w$ or $4x^{2}+4xy+7y^{2}=w$. Therefore,
if $w\equiv 1\pmod{24}$,

$N(-96,w)=|\\{(x,y)\in\mathbb{Z}:x^{2}+24y^{2}=w,\gcd(x,y)=1\\}|.$

In fact, as discussed in [8, pg. 84], a positive integer $w\equiv
1,5,7,11\pmod{24}$ can be represented by one and only one of the four
quadratic forms above. In the language of genera, this means that the four
quadratic forms lie in four different different genera.

In general, the number of positive, reduced, primitive quadratic forms of a
certain disciminant $-d$ is equal to the class number $h(-d)$. The quadratic
form $x^{2}+216y^{2}$ has discriminant $-864$ and there are twelve positive,
reduced, primitive quadratic forms with this discriminant. The total number of
solutions $N(-864,w)$ for all these quadratic forms can again be calculated
using [5, Lemma 3.25]. However, these quadratic forms do not lie in twelve
different genera, i.e., we can not separate them according to the congruence
classes for $w$, regardless of the choice of modulus. A number $n$ such that
the different positive, reduced, primitive quadratic forms with discriminant
$-4n$ lie in different genera, is an idoneal number or Euler convenient
number. Thus, $216$ is not idoneal.

Even though we can not obtain an explicit formula form $N_{2}(pm)$, in section
2.2 we are able to prove that, for $p\in\mathcal{P}$, $m\geq 1$ with $pm\equiv
1\pmod{24}$ and $p\nmid m$,

$N_{2}(pm)=2\cdot|\\{(a,b)\in\mathbb{Z}:a^{2}+216b^{2}=jm,\,\gcd(a,b)=1\\}|,$

where $j\in\\{1,4,8\\}$ is such that $x^{2}+216y^{2}=jp$ has primitive
solutions. This relation is enough to prove Theorem 1.3. In section 4 we give
a simple conjectured formula for $N_{2}(w)$.

The article is organized as follows. In section 2, we prove Theorem 1.3. In
subsection 2.1, we prove the theorem for small values of $p\in\mathcal{P}$
using modular forms and [18, Lemma 4.5]. We introduce all necessary notation
and discuss several hypothesis of the lemma for general primes $p\geqslant 5$.
Then, we show that the remaining hypotheses hold for $p\in\mathcal{P}$. Lemma
4.5 in [18] produces a bound and reduces the proofs to checking that the
congruences hold for integers up to the given bound. We performed the required
check. However, as $p$ grows, it becomes impossible to check the required
finite but large number of congruences. In subsection 2.2, we give a proof of
Theorem 1.3 (for all $p\in\mathcal{P}$) that uses only facts about quadratic
forms and determine the Dirichlet density of the set $\mathcal{P}$. The
results of section 3 are motivated by recent work of the second author [15].
Recall that Schur’s identity states that the number of partitions of $n$ into
parts congruent to $\pm 1$ modulo $6$ equals the number of $3$-regular
distinct partitions of $n$. We generalize the Schur reversal result of [15] to
all Euler-type identities, i.e., the number of partitions of $n$ with parts
parts from a set $S_{2}$ is equal to the number of distinct partition of $n$
with parts from a set $S_{1}$. We prove analytically and combinatorially that,
if $S_{1},S_{2}$ are subsets of integers for which an Euler-type identity
holds, then the number of distinct partitions with parts from $S_{2}$ weighted
by the parity of the number of parts equals the number of partitions with
parts from $S_{1}$ weighted by the parity of the number of parts. In section
4, we conjecture and exact formula for the number of primitive solutions for
$x^{2}+216y^{2}=m$ when $m\equiv 1\pmod{24}$. We end the section with
companion conjectures for the parity of the number of $3$-regular partitions
of $n$ with an even (respectively odd) number of parts.

##  2\. Proof of Theorem 1.3

We discovered Theorem 1.3 experimentally. Our initial guess was that the
theorem holds for

$\displaystyle p\in\mathcal{U}:=\\{29,59,79,103,223,$ $\displaystyle
227,241,251,269,293,337,419,443,487\\}.$

For these values of $p$, the theorem can be proved using modular forms and
[18, Lemma 4.5]. We present the proof here and follow with the proof for all
$p\in\mathcal{P}$. The latter does not use modular forms.

First we introduce some necessary background on partitions.

We denote by $p(n)$ the number of the partitions of $n$. For example, $p(5)=7$
since the partitions of $5$ are

$(5),\ (4,1),\ (3,2),\ (3,1,1),\ (2,2,1),\ (2,1,1,1),\ (1,1,1,1,1).$

The generating function of $p(n)$ is given by

$\sum_{n=0}^{\infty}p(n)\,q^{n}=\frac{1}{(q;q)_{\infty}}.$

Here and throughout $q$ is a complex number with $|q|<1$, and the symbol
$(a;q)_{\infty}$ denotes the infinite product

$(a;q)_{\infty}:=\prod_{n=0}^{\infty}(1-aq^{n}).$

The generating function for the sequence $b_{\ell}(n)$ is given by

$\sum_{n=0}^{\infty}b_{\ell}(n)q^{n}=\frac{(q^{\ell};q^{\ell})_{\infty}}{(q;q)_{\infty}}.$

###  2.1. Proof of Theorem 1.3 for $p\in\mathcal{U}$

Many proofs of congruences in the literature use [18, Lemma 4.5] and they
introduce all necessary notation. The reader can consult, for example, [19]
and [4]. For completeness and the convenience of the reader, we also define
all needed notation. Let $\Gamma:=SL(2,\mathbb{Z}),$ and define

$\Gamma_{\infty}:=\left\\{\left(\begin{array}[]{cc}1&b\\\
0&1\end{array}\right)\in\Gamma\right\\}.$

For a positive integer $N$, we define the congruence subgroup

$\Gamma_{0}(N):=\left\\{\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in\Gamma:c\equiv 0\pmod{N}\right\\}.$

If $M$ is a positive integer, we write $R(M)$ for the set of finite integer
sequences $r=(r_{\delta_{1}},r_{\delta_{2}},\ldots,r_{\delta_{k}})$, where
$1={\delta_{1}}<{\delta_{2}}<\cdots<{\delta_{k}}=M$ are the positive divisors
of $M$. For the remainder of the argument, we only consider positive divisors
of a given integer. Given a positive integer $m$, we denote by $S_{24m}$ the
set of invertible quadratic residues modulo $24m$ and, for fixed $0\leqslant
t\leqslant m-1$, we define

$P_{m,r}(t):=\left\\{ts+\frac{s-1}{24}\sum_{\delta\mid M}\delta r_{\delta}\mod
m:s\in S_{24m}\right\\},$

where by $x\mod m$ we mean the residue of $x$ modulo $m$.

Let $m,M$ and $N$ be positive integers. Moreover, let $t$ be an integer such
that $0\leqslant t\leqslant m-1$ and let $r=(r_{\delta})\in R(M)$. We set
$\kappa:=\gcd(1-m^{2},24)$ and write $\prod_{\delta\mid
M}\delta^{|r_{\delta}|}=:2^{s}\ell$, where $s$ is a non-negative integer and
$\ell$ is odd. Then, we say that the tuple
$(n,M,N,(r_{\delta}),t)\in\Delta^{*}$ if and only if all of the following six
conditions are satisfied.

  1. (1)

$p\mid m$, $p$ prime, implies $p\mid N$;

  2. (2)

$\delta\mid M$, $\delta\geqslant 1$ such that $r_{\delta}\neq 0$ implies
$\delta\mid mN$;

  3. (3)

$\kappa N\sum_{\delta\mid M}r_{\delta}\frac{mN}{\delta}\equiv 0\pmod{24}$;

  4. (4)

$\kappa N\sum_{\delta\mid M}r_{\delta}\equiv 0\pmod{8}$;

  5. (5)

$\displaystyle\frac{24m}{\gcd(\kappa(-24t-\sum_{\delta\mid M}\delta
r_{\delta}),24m)}\Big{|}N$;

  6. (6)

If $2\mid m$, then $(4\mid\kappa N\text{ and }8\mid Ns)$ or $(2\mid s\text{
and }8\mid N(1-j))$.

Finally, for $\gamma=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in\Gamma$, we define

$p_{m,r}(\gamma):=\min_{d\in\\{0,\ldots,m-1\\}}\frac{1}{24}\sum_{\delta\mid
M}r_{\delta}\frac{\gcd^{2}(\delta(a+\kappa dc),mc)}{\delta m}$

and

$p^{*}_{r}(\gamma):=\frac{1}{24}\sum_{\delta\mid
M}r_{\delta}\frac{\gcd^{2}(\delta,c)}{\delta}.$

With all the notation in place, we can now state [18, Lemma 4.5] that will
allow us to prove the congruences of Theorem 1.3 for $p\in\mathcal{U}$.

######  Lemma 2.1.

Let $u$ be a positive integer, $(m,M,N,t,r=(r_{\delta}))\in\Delta^{*}$,
$a=(a_{\delta})\in R(N)$. Let
$\\{\gamma_{1},\ldots,\gamma_{n}\\}\subset\Gamma$ be a complete set of
representatives of the double cosets in
$\Gamma_{0}(N)\backslash\Gamma/\Gamma_{\infty}$. Assume that
$p_{m,r}(\gamma_{i})+p^{*}_{s}(\gamma_{i})\geqslant 0$ for all $0\leqslant
i\leqslant n$. Let $t_{\min}:=\min_{t^{\prime}\in P_{m.r}(t)}t^{\prime}$ and

$\nu:=\frac{1}{24}\left(\left(\sum_{\delta\mid N}a_{\delta}+\sum_{\delta\mid
M}r_{\delta}\right)[\Gamma:\Gamma_{0}(N)]-\sum_{\delta\mid N}\delta
a_{\delta}\right)-\frac{1}{24m}\sum_{\delta\mid M}\delta
r_{\delta}-\frac{t_{\min}}{m}.$

Suppose

$\prod_{\delta\mid M}\prod_{n=1}^{\infty}(1-q^{\delta
n})^{r_{\delta}}=\sum_{n=0}^{\infty}c_{r}(n)q^{n}.$

If

$\sum_{n=0}^{\lfloor\nu\rfloor}c_{r}(mn+t^{\prime})q^{n}\equiv 0\pmod{u},\ \
\text{ for all }t^{\prime}\in P_{m,r}(t),$

then

$\sum_{n=0}^{\infty}c_{r}(mn+t^{\prime})q^{n}\equiv 0\pmod{u},\ \ \text{ for
all }t^{\prime}\in P_{m,r}(t).$

Recall that

$\sum_{n=0}^{\infty}b_{3}(n)q^{n}=\prod_{n=1}^{\infty}\frac{(1-q^{3n})}{(1-q^{n})}.$

Let

$\sum_{n=0}^{\infty}b(n)q^{n}:=\prod_{n=1}^{\infty}\frac{(1-q^{n})^{4}}{1-q^{3n}}$

In [13, pg. 287] it is shown that

$\sum_{n=0}^{\infty}b_{3}(2n)q^{n}\equiv\sum_{n=0}^{\infty}b(n)q^{n}\pmod{2}.$

We will prove below that

$b(p^{2}\,n+p\alpha-24_{p}^{-1})\equiv 0\pmod{2}$

where $p\in\mathcal{U}$ and $\alpha\neq\lfloor p/24\rfloor$ is a residue
modulo $p$.

To use the Lemma 2.1, we write

$\prod_{n=1}^{\infty}\frac{(1-q^{n})^{4}}{1-q^{3n}}=\prod_{\delta\mid
M}\prod_{n=1}^{\infty}(1-q^{\delta n})^{r_{\delta}}.$

Since we want to prove congruences of the form
$b(p^{2}\,n+p\alpha-24_{p}^{-1})\equiv 0\pmod{2}$ for certain primes $p$ and
integers $0\leqslant\alpha\leqslant p-1$, $p\neq\lfloor p/24\rfloor$, we have
$u=2$ and $m=p^{2}$. Then, $\kappa=24$. We let $M=3$, and
$(r_{1},r_{3})=(4,-1)$. Since $\prod_{\delta\mid M}\delta^{|r_{\delta}|}=3$,
we have $s=0$ and $\ell=3$. We take $N=3p$ and verify that for any
$t=p\alpha-24_{p}^{-1}$ as in the statement of Theorem 1.3 conditions 1.-6.
are satisfied.

Since

$[\Gamma:\Gamma_{0}(N)]=N\prod_{x\mid N}(1+x^{-1}),$

where the product is taken after all prime divisors of $N$, we have
$[\Gamma:\Gamma_{0}(N)]=4(p+1).$

Since $N$ is square free, by [21, Lemma 4.3], a complete set of
representatives for $\Gamma_{0}(N)\backslash\Gamma/\Gamma_{\infty}$ is given
by

$\mathcal{A}_{N}=\left\\{\left(\begin{array}[]{cc}1&0\\\
\delta&1\end{array}\right):\delta\mid N,\delta\geqslant 1\right\\}.$

Here, the positive divisors of $N=3p$ are $\\{1,3,p,3p\\}$.

We complete the proof of the congruences of Theorem 1.3 for $p\in\mathcal{U}$
case by case. For the discussion below we used Mathematica to calculate all
needed data. Let

$\mathcal{R}_{p}:=\\{p\alpha-24_{p}^{-1}:0\leqslant\alpha<p,\
\alpha\neq\lfloor p/24\rfloor\\}.$

In each case below, we verified that $p_{m,r}(\gamma)>0$ for all
$\gamma\in\mathcal{A}_{N}$ and thus we can take $a_{\delta}=0$ for all
$\delta\mid N$. We give more details in the case $p=29$, including the values
for $p_{m,r}(\gamma)$, and omit the details in the other cases.

Case $p=29$. We have $24_{p}^{-1}\equiv-6\pmod{29}$ and $\lfloor
p/24\rfloor=1$. We find that

$P(6)=\\{6,64,151,180,209,238,296,412,499,615,673,702,731,760\\}$

and

$P(3\cdot 29+6)=\\{93,122,267,325,354,383,441,470,528,557,586,644,789,818\\}.$

Thus,

$P(6)\cup P(3\cdot 29+6)=\mathcal{R}_{29}.$

Next we calculate $p_{m,r}(\gamma)$ for each $\gamma\in\mathcal{A}_{N}$. We
obtain

$\left\\{\frac{11}{60552},\frac{1}{20184},\frac{11}{60552},\frac{1}{20184}\right\\},$

where the terms are listed in increasing order of $\delta\mid N$ defining
$\gamma$.

Thus, for

$\displaystyle a:=(a_{1},a_{3},a_{5},a_{15})=(0,0,0,0)$

we have $p_{m,r}(\gamma)+p^{*}_{s}(\gamma)\geqslant 0$ for all
$\gamma\in\mathcal{A}_{N}.$

From Lemma 2.1, we obtain $\lfloor\nu\rfloor=14$ when $t=6$ and also when
$t=3\cdot 29+6=93$.

It is easily verified that for each $t\in\mathcal{R}_{29}$, we have

$b(29^{2}n+t)\equiv 0\pmod{2}\text{ for }0\leqslant n\leqslant 14.$

Thus, by Lemma 2.1, for each $t\in\mathcal{R}_{29}$, we have

$b(29^{2}n+t)\equiv 0\pmod{2}\text{ for all }n\geqslant 0.$

For the other primes in $\mathcal{U}$, we list the relevant data needed for
Lemma 2.1 as well as the obtained bound in the table below. In the table, $A$
is the smallest positive integer not equal to $\lfloor p/24\rfloor$ such that
$P(-24_{p}^{-1})\cup P(Ap-24_{p}^{-1})=\mathcal{R}_{p}.$ In each case, we used
Mathematica to verify that $p_{m,r}(\gamma)>0$ for each
$\gamma\in\mathcal{A}_{N}$, and we used
$a:=(a_{1},a_{3},a_{5},a_{15})=(0,0,0,0)$ to calculate $\lfloor\nu\rfloor$.

$\begin{array}[]{c||c|c|c|c}\quad p&\quad-24_{p}^{-1}&\quad\lfloor
p/24\rfloor&\quad A&\quad\lfloor\nu\rfloor\\\ \hline\cr\hline\cr
59&27&2&1&29\\\ \hline\cr 79&23&3&1&39\\\ \hline\cr 103&30&4&1&51\\\ \hline\cr
223&65&9&3&111\\\ \hline\cr 227&104&9&1&113\\\ \hline\cr 241&10&10&3&120\\\
\hline\cr 251&115&10&1&125\\\ \hline\cr 269&56&11&1&134\\\ \hline\cr
293&61&12&2&146\\\ \hline\cr 337&14&14&3&168\\\ \hline\cr 419&192&17&1&209\\\
\hline\cr 443&203&18&1&221\\\ \hline\cr 487&142&20&1&243\end{array}$

In each case, we verified the congruences up to the bound $\lfloor\nu\rfloor$
given by Lemma 2.1. Hence, for $p\in\mathcal{U}$, the congruences hold for all
$n\geqslant 0$. We needed to compute $b_{3}(m)$ modulo $2$ for all $m\leqslant
2(487^{2}\cdot 243+486)=115\,265\,106$. Since this was computationally
expensive, we had to optimize the algorithm generating $b_{3}(m)$ modulo $2$
several times.

###  2.2. Proof of Theorem 1.3 for all $p\in\mathcal{P}$

From [13], we have

$\displaystyle\sum_{n=0}^{\infty}b_{3}(2n)q^{n}$
$\displaystyle\equiv\sum_{n\in\mathbb{Z}}q^{\frac{n}{2}(3n-1)^{2}}\left(1+\sum_{n\in\mathbb{Z}}q^{(3n-1)^{2}}\right)$
$\displaystyle\equiv\sum_{n\in\mathbb{Z}}q^{\frac{n}{2}(3n-1)}+\sum_{n,m\in\mathbb{Z}}q^{\frac{n}{2}(3n-1)+(3m-1)^{2}}$
$\displaystyle=:\sum_{r=0}^{\infty}a(r)q^{r},$

where $a(r)$ is the number of representations of $r$ as the sum of a
generalized pentagonal number and a the square of a an integer $y$ such that
$y\not\equiv 0\pmod{3}$ if $y\neq 0$. Thus,

$a(r)=|\\{(n,m)\in\mathbb{Z}^{2};\
\frac{n}{2}(3n-1)+(3m-1)^{2}=r\\}|+\delta_{r=\frac{n}{2}(3n-1)}.$

Here, $\delta_{\rho}$ is the Kronecker delta function which equals $1$ if
statement $\rho$ is true and $0$ otherwise. We can rewrite $a(r)$ as

$a(r)=|\\{(x,y)\in\mathbb{N}^{2};\ x^{2}+24y^{2}=24r+1,y=0\text{ or }3\nmid
y.\\}|$

Our goal is to prove that for $p\in\mathcal{P}$, $n\geqslant 0$,
$0\leqslant\alpha<p$, $\alpha\neq\lfloor p/24\rfloor$, we have
$a(p^{2}\,n+p\,\alpha-24_{p}^{-1})\equiv 0\pmod{2}.$

####  2.2.1. Preliminary lemmas

In this subsection, we prove several lemmas which will be key ingredients in
the proof of the theorem.

######  Lemma 2.2.

For any prime $p\geqslant 5$ and $0\leqslant\alpha\leqslant p-1$,
$\alpha\neq\lfloor\frac{p}{24}\rfloor$, we have $p^{2}\nmid
24(p\alpha-24^{-1}_{p})+1$.

###### Proof.

Recall that $-24^{-1}_{p}$ as an integer between $1$ and $p-1$. Then,
$24\cdot(-24^{-1}_{p})=tp-1$ with $1\leqslant t\leqslant 23$. Thus,
$24(p\alpha-24^{-1}_{p})+1=24p\alpha+tp$.

Suppose $p^{2}\mid 24(p\alpha-24^{-1}_{p})+1$. Then
$24(p\alpha-24^{-1}_{p})+1=kp^{2}$ for some $k\in\mathbb{Z}$. Since
$p^{2}\equiv 1\pmod{24}$, we must have $k\equiv 1\pmod{24}$. Since
$1\leqslant\alpha<p$ and $1\leqslant-24^{-1}_{p}\leqslant p-1$, we have
$1\leqslant p\alpha-24^{-1}_{p}\leqslant p^{2}-1$ and we must have $k=1$.
Then, $24(p\alpha-24^{-1}_{p})+1=p^{2}$, and thus $24\alpha+t=p$. Finally,
since $1\leqslant t\leqslant 23$, we have $\alpha=\lfloor\frac{p}{24}\rfloor$
which is a contradiction. ∎

######  Corollary 2.3.

For any prime $p\geqslant 5$ and $0\leqslant\alpha<p$,
$\alpha\neq\lfloor\frac{p}{24}\rfloor$, and $s=p^{2}n+p\alpha-24^{-1}_{p}$.
Then, $24s+1=pm$, for some positive integer $m$ such that $p\nmid m$.

###### Proof.

We have $24s+1=24p^{2}n+24p\alpha-24^{-1}_{p}\cdot 24+1$ and the result
follows immediately form Lemma 2.2. ∎

Next, for $m$ coprime to $6$ and $j\in\\{1,4,8\\}$, we consider the
possibilities for $m\pmod{24}$ when

(2) $x^{2}+216y^{2}=jm$

has primitive solutions.

######  Lemma 2.4.

Let $m$ be a positive integer coprime to $6$. If equation 2 with $j=1$ has
primitive solutions, then $m\equiv 1\pmod{24}$.

###### Proof.

Let $m>0$, $\gcd(m,6)=1$ and let $x,y\in\mathbb{Z}$ with $\gcd(x,y)=1$ be such
that $x^{2}+216y^{2}=m$, Then, $2,3\nmid x$ and thus $x^{2}\equiv 1\pmod{24}$.
It follows that $m\equiv 1\pmod{24}$. ∎

######  Lemma 2.5.

Let $m$ be a positive integer coprime to $6$. If equation 2 with $j=4$ has
primitive solutions, then $m\equiv 7\pmod{24}$.

###### Proof.

Let $m>0$, $\gcd(m,6)=1$ and let $x,y\in\mathbb{Z}$ with $\gcd(x,y)=1$ be such
that $x^{2}+216y^{2}=4m$. Then $x=2z$ for some $z\in\mathbb{Z}$ and we have
$z^{2}+54y^{2}=m$. Therefore, $2,3\nmid z$ and thus $z^{2}\equiv 1\pmod{24}$.
Since $\gcd(x,y)=1$, it follows that $2\nmid y$ and $y^{2}\equiv
1,9\pmod{24}$. Then $m\equiv 1+6y^{2}\equiv 7\pmod{24}$. ∎

######  Lemma 2.6.

Let $m$ be a positive integer coprime to $6$. If equation 2 with $j=8$ has
primitive solutions, then $m\equiv 5,11\pmod{24}$. Moreover, if $(x,y)$ is a
primitive solution for (2) with $j=8$, then

  * (i)

if $m\equiv 5\pmod{24}$, then $\textrm{val}_{2}(x)=2$;

  * (ii)

f $m\equiv 11\pmod{24}$, then $\textrm{val}_{2}(x)\geqslant 3$.

###### Proof.

Let $m>0$, $\gcd(m,6)=1$ and suppose $x,y\in\mathbb{Z}$ are such that
$\gcd(x,y)=1$ and $x^{2}+216y^{2}=8m$. Then, $x=4z$ for some $z\in\mathbb{Z}$
and

$2z^{2}+27y^{2}=m.$

Since $\gcd(m,6)=1$, it follows that $2\nmid y$ and $3\nmid z$. Then, one can
check that $y^{2}\equiv 1,9\pmod{24}$ and $z^{2}\equiv 1,4,16\pmod{24}$. A
case by case verification shows that if $z^{2}\equiv 1\pmod{24}$, and
$y^{2}\equiv 1,9\pmod{24}$, then $m\equiv 5\pmod{24}$, and if $z^{2}\equiv
4,16\pmod{24}$, and $y^{2}\equiv 1,9\pmod{24}$, then $m\equiv 11\pmod{24}$.
Thus, if $m\equiv 5\pmod{24}$, then $\textrm{val}_{2}(z)=0$ and thus
$\textrm{val}_{2}(x)=2$, and if $m\equiv 11\pmod{24}$, then
$\textrm{val}_{2}(z)\geqslant 1$ and thus $\textrm{val}_{2}(x)\geqslant 3$. ∎

In particular, the previous three lemmas show that, if $p\in\mathcal{P}$, then

$\displaystyle j=1$ $\displaystyle\implies p\equiv 1\pmod{24}$ $\displaystyle
j=4$ $\displaystyle\implies p\equiv 7\pmod{24}$ $\displaystyle j=8$
$\displaystyle\implies p\equiv 5,11\pmod{24},$

which was noted in Remark 1.

######  Lemma 2.7.

Let $p\in\mathcal{P}$ and let $m$ be a positive integer coprime to $p$ such
that $pm\equiv 1\pmod{24}$. Let $j\in\\{1,4,8\\}$ such that
$x^{2}+216y^{2}=jp$ has primitive solutions. Define

$\displaystyle X_{p,m}$
$\displaystyle:=\\{(x,y)\in\mathbb{Z}^{2}:x^{2}+216y^{2}=pm,\gcd(x,y)=1\\}$
$\displaystyle A_{m}$
$\displaystyle:=\\{(a,b)\in\mathbb{Z}^{2}:a^{2}+216b^{2}=jm,\gcd(a,b)=1\\}.$

Then $|X_{p,m}|=2|A_{m}|$.

###### Proof.

Let $p\in\mathcal{P}$ and $j\in\\{1,4,8\\}$ be such that $x^{2}+216y^{2}=jp$
has primitive solutions and fix $x_{1},y_{1}\in\mathbb{Z}$ with
$\gcd(x_{1},y_{1})=1$ satisfying $x_{1}^{2}+216y_{1}^{2}=jp$. Let $m>1$ be
such that $p\nmid m$ and $pm\equiv 1\pmod{24}$. We define a mapping
$f:X_{p,m}\to A_{m}$ by $f(x,y)=(a,b)$ as follows. Start with $(x,y)\in
X_{p,m}$.

  * (i)

If $x_{1}x-216y_{1}y\equiv 0\pmod{p}$, one can show that $x_{1}y+y_{1}x\equiv
0\pmod{p}$ and we define

$\displaystyle a$ $\displaystyle:=\frac{x_{1}x-216y_{1}y}{p},\qquad
b:=\frac{x_{1}y+y_{1}x}{p}.$

  * (ii)

If $x_{1}x+216y_{1}y\equiv 0\pmod{p}$, one can show that $x_{1}y-y_{1}x\equiv
0\pmod{p}$ and we define

$\displaystyle a$ $\displaystyle:=\frac{x_{1}x+216y_{1}y}{p},\qquad
b:=\frac{x_{1}y-y_{1}x}{p}.$

Then, in both cases

$a^{2}+216b^{2}=\frac{(x_{1}^{2}+216y_{1}^{2})(x^{2}+216y^{2})}{p^{2}}=jm.$

To see that $\gcd(a,b)=1$, express $x$ and $y$ in terms of $a$ and $b$. In
case (i) we have

(3) $x=\frac{x_{1}a+216y_{1}b}{j},\qquad y=\frac{bx_{1}-ay_{1}}{j},$

and in case (ii) we have

(4) $x=\frac{x_{1}a-216y_{1}b}{j},\qquad y=\frac{bx_{1}+ay_{1}}{j}.$

Since $pm$ is odd, $x$ must be odd. If $j=4,8$, then $x_{1}$ is even and since
$\gcd(x_{1},y_{1})=1$, it follows that $y_{1}$ is odd. Therefore, $a$ is even
and $b$ is odd, and from (3) and (4), we have that $\gcd(a,b)\mid\gcd(x,y)$.
Hence $\gcd(a,b)=1$ and $(a,b)\in A_{m}$.

Next, we show that $f$ is surjective and two-to-one. Start with $(a,b)\in
A_{m}$ and define $x,y$ by the formulas given in (3) and $\tilde{x},\tilde{y}$
by the formulas given in (4). It is straight forward to verify that

$\displaystyle x^{2}+216y^{2}$ $\displaystyle=pm$
$\displaystyle\tilde{x}^{2}+216\tilde{y}^{2}$ $\displaystyle=pm.$

As before, if $j=4$, then $2\mid a,x_{1}$. If $j=8$, then $4\mid a,x_{1}$.
Thus, in all cases $x,\tilde{x}\in\mathbb{Z}$.

Since $\gcd(a,b)=\gcd(x_{1},y_{1})=1$, it follows that $2\nmid b,y_{1}$. Then,
if $j=4$, $y,\tilde{y}\in\mathbb{Z}$.

If $j=8$, by Lemma 2.6, $p\equiv 5,11\pmod{24}$. If $p\equiv 5\pmod{24}$,
$\textrm{val}_{2}(a)=\textrm{val}_{2}(x_{1})=2$. As above, $2\nmid b,y_{1}$
and thus $y,\tilde{y}\in\mathbb{Z}$. If $p\equiv 5\pmod{24}$, we have
$\textrm{val}_{2}(a)$, $\textrm{val}_{2}(x_{1})\geqslant 3$ and thus
$y,\tilde{y}\in\mathbb{Z}$.

A similar discussion as above for $a,b$, shows that
$\gcd(x,y)=\gcd(\tilde{x},\tilde{y})=1$. Thus, $(x,y),(\tilde{x},\tilde{y})\in
X_{p,m}$.

Moreover, one can check that

(5) $\displaystyle x_{1}x-216y_{1}y\equiv 0\pmod{p}$ (6) $\displaystyle
x_{1}\tilde{x}+216y_{1}\tilde{y}\equiv 0\pmod{p}.$

Hence, $f^{-1}(a,b)=\\{(x,y),(\tilde{x},\tilde{y})\\}$.

It remains to show that $(x,y)\neq(\tilde{x},\tilde{y})$. Suppose
$(x,y)=(\tilde{x},\tilde{y})$. Then each pair $(x,y),(\tilde{x},\tilde{y})$
satisfies both congruences (5) and (6). As a consequence, $x_{1}x\equiv
0\pmod{p}$ which contradicts $\gcd(x,y)=\gcd(x_{1},y_{1})=1$.

This shows that $f$ is surjective and two-to-one. ∎

######  Corollary 2.8.

With $p,m,j$ and $X_{p,m}$ as in Lemma 2.7, $|X_{p,m}|\equiv 0\pmod{8}$ if
$m>1$. If $m=1$, then $|X_{p,m}|=4$.

###### Proof.

If $m=1$, then $p\equiv 1\pmod{24}$ and therefore $j=1$. Then $(1,0)$ and
$(-1,0)$ are the only elements of $A_{m}$ and $|X_{p,m}|=2|A_{m}|=4$. If $m>1$
and $(a,b)\in A_{m}$, it follows that $a,b\neq 0$. Then, $(a,b)\in A_{m}$ if
and only if $(-a,-b),(-a,b),(a,-b)\in A_{m}$. Hence, $|A_{m}|\equiv
0\pmod{4}$, and by Lemma 2.6, $|X_{p,m}|\equiv 0\pmod{8}$. ∎

####  2.2.2. Proof of Theorem 1.3

We introduce the following notation.

$\displaystyle M_{1}(w)$
$\displaystyle:=|\\{(x,y)\in\mathbb{Z}^{2}:x^{2}+24y^{2}=w\\}|$ $\displaystyle
M_{2}(w)$ $\displaystyle:=|\\{(x,y)\in\mathbb{Z}^{2}:x^{2}+24\cdot
9y^{2}=w\\}|$ $\displaystyle N_{1}(w)$ $\displaystyle:=|\\{(x,y)\in
M_{1}(w):{\gcd(x,y)=1}\\}|$ $\displaystyle N_{2}(w)$
$\displaystyle:=|\\{(x,y)\in M_{1}(w):\gcd(x,y)=1\\}|.$

Note that if $u$ is square free, then $M_{i}(u)=N_{i}(u)$.

From [8, pg. 84], if $w\equiv 1\pmod{24}$, we have

(7) $M_{1}(w)=2\sum_{d\mid w}\Big{(}\frac{-6}{d}\Big{)},$

where the sum is over all positive divisors of $w$ and
$\big{(}\frac{-6}{p}\big{)}$ is the Jacobi symbol. As discussed in the
introduction, since $24$ is an idolean number, if $w\equiv 1\pmod{24}$, we
have

(8) $N_{1}(w)=N(-96,w)=2\prod_{p\mid
w}\left(1+\Big{(}\frac{-6}{p}\Big{)}\right),$

where the product is over all prime divisors of $w$.

In particular, if $p\in\mathcal{P}$ and $p\equiv 1\pmod{24}$, then
$M_{1}(p)=N_{1}(p)=4$.

Recall that $s=p^{2}n+p\alpha-24^{-1}_{p}$ and $24s+1=pm$ with $p\nmid m$. If
$(x,y)\in\mathbb{Z}^{2}$, is a solution for $x^{2}+24y^{2}=24s+1$, then $x\neq
0$ and, since $24s+1=pm$ is not a perfect square, we also have $y\neq 0$.
Then, as in the proof of Corollary 2.8, $M_{i}(pm)\equiv N_{i}(pm)\equiv
0\pmod{4}$.

Since $pm$ is not a perfect square and $pm\equiv 1\pmod{24}$, to prove that
$b_{3}(2s)\equiv 0\pmod{2}$, we need to show that

$b_{3}(2s)\equiv
a(s)\equiv\frac{1}{4}\Big{(}M_{1}(24s+1)-M_{2}(24s+1)\Big{)}\pmod{2}.$

Hence, we need to show that

$M_{1}(pm)-M_{2}(pm)\equiv 0\pmod{8}.$

If $m\neq 1$, it follows from (8) that $N_{1}(pm)\equiv 0\pmod{8}$. As
previously mentioned, if $u$ is square free then $N_{i}(u)=M_{i}(u)$. If
$u=gh^{2}$ with $g$ square free, then $M_{i}(u)=\sum_{d|h}N_{i}(gd^{2})$.

Case I: If $m$ is not a perfect square, then $pm=pgh^{2}$ for some positive
integers $h,g$ with $g$ square free. Then

$M_{i}(pm)=M_{i}(pgh^{2})=\sum_{d|h}N_{i}(pgd^{2}).$

Since $pm=pgh^{2}\equiv 1\pmod{24}$, $\gcd(h,6)=1$ and $h^{2}\equiv
1\pmod{24}$. Then $pg\equiv 1\pmod{24}$ and thus $pgd^{2}\equiv 1\pmod{24}$
for all $d\mod h$. By Corollary 2.8, $N_{i}(pgd^{2})\equiv 0\pmod{8}$ for all
$d\mid h$ and thus $M_{i}(pm)\equiv 0\pmod{8}$.

Case II: If $m$ is a perfect square, since $\gcd(m,6)=1$, we have $m\equiv
1\pmod{24}$. Then $p\equiv 1\pmod{24}$ and $N_{1}(p)=4$. We have

$M_{i}(pm)=\sum_{\begin{subarray}{c}d|h\\\
d>1\end{subarray}}N_{i}(pd^{2})+N_{i}(p).$

By Corollary 2.8, $N_{2}(pd^{2})\equiv 0\pmod{8}$ for all $d\mid m$, $d>1$,
and it follows that $M_{i}(pm)\equiv N_{i}(p)\pmod{8}$.

Since $p\in\mathcal{P}$, $N_{2}(p)>0$. Since $N_{2}(p)\leqslant N_{1}(p)$ and
$N_{2}(p)\equiv 0\pmod{4}$, we have $N_{1}(p)=N_{2}(p)=4.$ Therefore
$M_{1}(pm)-M_{2}(pm)\equiv 0\pmod{8}$. This concludes the proof.

####  2.2.3. The Dirichlet density of $\mathcal{P}$

In this subsection we determine the Dirichlet density of the set
$\mathcal{P}$. Since the density is positive, Theorem 1.3 holds for infinitely
many primes. Moreover, we show that Theorem 1.3 holds for infinitely many
primes congruent to $k\pmod{24}$ for each $k\in\\{1,5,7,11\\}$.

######  Definition 1.

Let $\Sigma\subset\mathbb{N}$ be a subset of prime numbers. Then the Dirichlet
density, $\delta(\Sigma)$, of $\Sigma$ is defined as

$\delta(\Sigma):=\lim_{s\to
1^{+}}\frac{\sum_{p\in\Sigma}p^{-s}}{\sum_{p}p^{-s}},$

where the sum in the denominator is over all primes.

Using basic facts about the Riemann zeta function, it can be shown that

$\delta(\Sigma)=\lim_{s\to 1^{+}}\frac{\sum_{p\in\Sigma}p^{-s}}{-\log(s-1)}.$

For $k\in\\{1,5,7,11\\}$, let $\mathcal{P}_{k}=\\{p\in\mathcal{P}:p\equiv
k\pmod{24}\\}.$

######  Proposition 2.9.

The Dirichlet density $\delta(\mathcal{P})=\frac{1}{6}.$ Furthermore
$delta(\mathcal{P}_{k}))=\frac{1}{24}$ for $k\in\\{1,5,7,11\\}$.

###### Proof.

By Lemma 2.4,

$\mathcal{P}_{1}=\\{p:\text{there exist $x,y\in\mathbb{Z}$ such that
$x^{2}+216y^{2}=p$}\\}.$

By [5, Theorem 9.12] we have

$\delta(\mathcal{P}_{1})=\frac{1}{2h(-864)}=\frac{1}{24}.$

If there are $x,y\in\mathbb{Z}$ such that $\gcd(x,y)=1$ and
$x^{2}+216y^{2}=4p$, then $x=2z$ for some $z\in\mathbb{Z}$, and
$z^{2}+54y^{2}=p$. However if $(z,y)$ is a solution to $z^{2}+54y^{2}=p$, then
indeed $(x,y)=(2z,y)$ is a solution to $x^{2}+216y^{2}=4p$ but we might have
$\gcd(x,y)\neq 1$. It might be that $2|y$, in this case
$z^{2}+216(y/2)^{2}=p$. Therefore,

$\displaystyle\mathcal{P}_{7}=$ $\displaystyle\\{p:\text{there exist
$z,y\in\mathbb{Z}$ such that $z^{2}+54y^{2}=p$, $2\nmid y$}\\}$
$\displaystyle=$ $\displaystyle\\{p:\text{there exist $z,y\in\mathbb{Z}$ such
that $z^{2}+54y^{2}=p$}\\}$ $\displaystyle-\\{p:\text{there exist
$x,y\in\mathbb{Z}$ such that $x^{2}+216y^{2}=p$}\\}$

and by [5, Theorem 9.12] we have

$\delta(\mathcal{P}_{7})=\frac{1}{2h(-216)}-\frac{1}{2h(-864)}=\frac{1}{24}.$

Similarly,

$\mathcal{P}_{5}\cup\mathcal{P}_{11}=\\{p:\text{there exist $z,y\in\mathbb{Z}$
such that $2z^{2}+27y^{2}=p$}\\},$

and by [5, Theorem 9.12] we have

$\delta(\mathcal{P}_{5}\cup\mathcal{P}_{11})=\frac{1}{2h(-216)}=\frac{1}{12}.$

If $p\equiv 11\pmod{24}$ and $z,y$ satisfy $2z^{2}+27y^{2}=p$, then $z$ must
be even. Hence,

$\mathcal{P}_{11}=\\{p:\text{there exist $z,y\in\mathbb{Z}$ such that
$8z^{2}+27y^{2}=p$}\\},$

and by [5, Theorem 9.12] we have

$\delta(\mathcal{P}_{11})=\frac{1}{2h(-864)}=\frac{1}{24}.$

This implies that

$\delta(\mathcal{P}_{5})=\delta(\mathcal{P}_{5}\cup\mathcal{P}_{11})-\delta(\mathcal{P}_{11})=\frac{1}{12}-\frac{1}{24}=\frac{1}{24}.$

Therefore,

$\delta(\mathcal{P})=\delta(\mathcal{P}_{1})+\delta(\mathcal{P}_{7})+\delta(\mathcal{P}_{5}\cup\mathcal{P}_{11})=\frac{1}{24}+\frac{1}{24}+\frac{1}{12}=\frac{1}{6}.$

∎

Hence, for each $k\in\\{1,5,7,11\\}$, there are infinitely many primes
congruent to $k$ modulo $24$ satisfying the congruences of Theorem 1.3.

We conclude this section by showing that for $p\in\mathcal{P}$ we have
$b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv 1\pmod{2}$, which was noted in
Remark 2.

######  Proposition 2.10.

A prime $p$ belongs to $\mathcal{P}$ if and only if the Diophantine equation
$x^{2}+216y^{2}=p^{2}$ has primitive solutions.

###### Proof.

If $(x_{1},y_{1})$ is a primitive solution for $x^{2}+216y^{2}=jp$, where
$j=1,4,8$, an easy calculation shows that

$\displaystyle x^{\prime}$ $\displaystyle:=\frac{x_{1}^{2}-216y_{1}^{2}}{j}$
$\displaystyle y^{\prime}$ $\displaystyle:=\frac{2x_{1}y_{1}}{j}$

is a primitive solution for $x^{2}+216y^{2}=p^{2}$.

Conversely, let $x,y$ be coprime integers such that $x^{2}+216y^{2}=p^{2}$.
Then the pair $(X,Y):=(x,3y)$ is such that $X^{2}+24Y^{2}=p^{2}$ and
$\gcd(X,Y)=1$.

Suppose $p\equiv 1\pmod{24}$. Since $p\equiv 1\pmod{24}$, we have that
$N_{1}(p)=4$. So, there exist $a,b\in\mathbb{Z}$ such that $a^{2}+24b^{2}=p$
and $\gcd(a,b)=1$. This implies that the pair $(u,v)$ defined by

$(u+2\sqrt{-6}v):=(a+2\sqrt{-6}b)^{2}=a^{2}-24b^{2}+2\sqrt{-6}(2ab)$

is such that $u^{2}+24v^{2}=p^{2}$. Since $\gcd(a,b)=1$ and $\gcd(a,6)=1$, it
follows that $\gcd(u,v)=1.$ Then, $(x,3y)=(X,Y)\in\\{(\pm u,\pm v)\\}.$ Since
$v=2ab$, we must have $ab\equiv 0\pmod{3}$. Since $3\nmid a$, we have $3|b$.
Then, we can write $b=3c$ for some $c\in\mathbb{Z}$. This implies
$a^{2}+216c^{2}=p$ and $(a,c)$ is a primitive solution for $x^{2}+216y^{2}=jp$
with $j=1$.

Next, we consider the case $p\equiv 5,7,11\pmod{24}$. For $\gcd(w,6)=1$, the
number of primitive representations of $w$ by a positive, reduced, primitive
quadratic form of discriminant $-24$ is

$N(-24,w)=2\prod_{p|w}\left(1+\left(\frac{-6}{p}\right)\right).$

Since $h(-24)=2$, we have two quadratic forms of discriminant $-24$:
$x^{2}+6y^{2}$ and $2x^{2}+3y^{2}$. Therefore,

$\displaystyle N(-24,w)=$
$\displaystyle|\\{(x,y)\in\mathbb{Z}:x^{2}+6y^{2}=w,\gcd(x,y)=1\\}|$
$\displaystyle+|\\{(x,y)\in\mathbb{Z}:2x^{2}+3y^{2}=w,\gcd(x,y)=1\\}|.$

We define

$\displaystyle N_{3}(w)$
$\displaystyle:=|\\{(x,y)\in\mathbb{Z}:x^{2}+6y^{2}=w,\gcd(x,y)=1\\}|$
$\displaystyle N_{4}(w)$
$\displaystyle:=|\\{(x,y)\in\mathbb{Z}:2x^{2}+3y^{2}=w,\gcd(x,y)=1\\}|.$

We note that $6$ is an idoneal number. If $w\equiv 1,7\pmod{24}$ then only
$x^{2}+6y^{2}=w$ has solutions and so $N(-24,w)=N_{3}(w)$, and if $w\equiv
5,11\pmod{24}$ then only $2x^{2}+3y^{2}=w$ has solutions and so
$N(-24,w)=N_{4}(w)$.

First let $p\equiv 7\pmod{24}$. We have $N_{3}(p)=4$. Hence, there exist
$a,b\in\mathbb{Z}$ such that $a^{2}+6b^{2}=p$ and $\gcd(a,b)=1$. This implies
that the pair $(u,v)$ defined by

$u+2\sqrt{-6}v:=(a+\sqrt{-6}b)^{2}=a^{2}-6b^{2}+2\sqrt{-6}ab$

is such that $u^{2}+24v^{2}=p^{2}$. Moreover, $\gcd(u,v)=1$. Then,
$(x,3y)=(X,Y)\in\\{(\pm u,\pm v)\\}.$ Since $v=ab$ we must have $ab\equiv
0\pmod{3}$. Since $3\nmid a$ we have $3|b$. So we can write $b=3c$ for some
$c\in\mathbb{Z}$. This implies $a^{2}+54c^{2}=p$ or, equivalently,
$(2a)^{2}+216c^{2}=4p$. Next note that $2\nmid c$ because of $p\equiv
7\pmod{24}$, which implies $\gcd(2a,c)=1$. Hence, $(2a,c)$ is a primitive
solution for $x^{2}+216y^{2}=jp$ with $j=4$.

Finally, let $p\equiv 5,11\pmod{24}$. We have $N_{4}(p)=4$. Hence, there exist
$a,b\in\mathbb{Z}$ such that $2a^{2}+3b^{2}=p$ and $\gcd(a,b)=1$. This implies
that the pair $(u,v)$ defined by

$u+2\sqrt{-6}v:=(\sqrt{2}a+\sqrt{-3}b)^{2}=2a^{2}-3b^{2}+2\sqrt{-6}ab$

is such that $u^{2}+24v^{2}=p^{2}$. Moreover, $\gcd(u,v)=1$. Then,
$(x,3y)=(X,Y)\in\\{(\pm u,\pm v)\\}.$ Since $v=ab$ we must have $ab\equiv
0\pmod{3}$. Since $3\nmid a$ we have $3|b$. So we can write $b=3c$ for some
$c\in\mathbb{Z}$. This implies $2a^{2}+27c^{2}=p$ or, equivalently,
$(4a)^{2}+216c^{2}=8p$. Hence, $(4a,c)$ is a primitive solution for
$x^{2}+216y^{2}=jp$ with $j=8$.

∎

######  Corollary 2.11.

If $p\in\mathcal{P}$, then $b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv
1\pmod{2}$.

###### Proof.

To show that for $p\in\mathcal{P}$ we have
$b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv 1\pmod{2}$, we need to show that
$x^{2}+24y^{2}=p^{2}$ has an odd number of solutions $(x,y)\in\mathbb{N}^{2}$
such that $y=0$ or $3\nmid y$. If $p\in\mathcal{P}$, then $-6$ is a quadratic
residue modulo $p$ and from (7) we have $M_{1}(p^{2})=6$. Of the six solution
for $x^{2}+24y^{2}=p^{2}$, there are only two solutions in $\mathbb{N}^{2}$,
one of which is $(p,0)$. By Proposition 2.10, the other solution $(x,y)$
satisfies $y\neq 0$ and $3\mid y$. It follows that
$a\left(\frac{p^{2}-1}{24}\right)=1$ and thus
$b_{3}\left(\frac{p^{2}-1}{12}\right)\equiv 1\pmod{2}$. ∎

##  3\. Reversals of Euler type theorems

If $S$ is a subset of positive integers, we denote by $p(S;n)$ the number of
partitions of $n$ with parts from $S$. If $k\geqslant 2$, we denote by
$q_{k}(S;n)$ the number of partitions of $n$ with parts from $S$, each part
occurring at most $k-1$ times. If $k=2$, we write $q(S;n)$ for $q_{2}(S;n)$.
Thus, $q(S;n)$ is the number of distinct partitions of $n$ with parts from
$S$.

A pair $(S_{1},S_{2})$ of subsets of positive integers is called an Euler pair
if $q(S_{1};n)=p(S_{2};n)$ for all $n\geqslant 0$. Andrews [1] proved that
$(S_{1},S_{2})$ is an Euler pair if and only if $2S_{1}\subseteq S_{1}$ and
$S_{2}=S_{1}\setminus 2S_{1}$.

This notion has been generalized by Subbarao [20]. Given $k\geqslant 2$, the
pair $(S_{1},S_{2})$ is called an Euler pair of order $k$ if
$q_{k}(S_{1};n)=p(S_{2};n)$ for all $n\geqslant 0$. Subbarao proved that
$(S_{1},S_{2})$ is an Euler pair of order $k$ if and only if $kS_{1}\subseteq
S_{1}$ and $S_{2}=S_{1}\setminus kS_{1}$.

######  Example 1 (Subbarao [20]).

Let

$\displaystyle S_{1}$ $\displaystyle=\\{m\in\mathbb{N}:\ m\equiv
1\pmod{2}\\};$ $\displaystyle S_{2}$ $\displaystyle=\\{m\in\mathbb{N}:\
m\equiv\pm 1\pmod{6}\\}.$

Then $(S_{1},S_{2})$ is an Euler pair of order 3.

We note that Glaisher’s bijection used to prove Euler’s identity
$q(\mathbb{N},n)=p(2\mathbb{N}-1,n)$ can be generalized to prove the
corresponding partition identity for any Euler pair of order $r$.

Next, we consider a reversal of the partition identity given by the Euler pair
$(S_{1},S_{2})$. We denote by $p_{e}(S;n)$, respectively $p_{o}(S;n)$, the
number of partitions of $n$ with parts in $S$ and an even, respectively odd,
number of parts. We use analogous notation for other partition numbers.

######  Theorem 3.1.

Let $(S_{1},S_{2})$ be an Euler pair. For $n\geqslant 0$, we have

$q_{e}(S_{2};n)-q_{o}(S_{2};n)=p_{e}(S_{1};n)-p_{o}(S_{1},n).$

###### Analytic proof.

We have

$\displaystyle\sum_{n\geqslant 0}(p_{e}(S_{1};n)-p_{o}(S_{1},n))q^{n}$
$\displaystyle=\prod_{j\in S_{1}}\frac{1}{1+q^{j}}=\prod_{j\in
S_{1}}\frac{1-q^{j}}{1-q^{2j}}$ $\displaystyle=\prod_{j\in
S_{1}}(1-q^{j})\prod_{j\in 2S_{1}}\frac{1}{1-q^{j}}$
$\displaystyle=\prod_{j\in S_{1}}(1-q^{j})\prod_{j\in S_{1}\setminus
S_{2}}\frac{1}{1-q^{j}}$ $\displaystyle=\prod_{j\in S_{2}}(1-q^{j})$

This completes the proof. ∎

###### Combinatorial proof.

Our combinatorial proof of this result is similar to that in [10]. Since
$(S_{1},S_{2})$ is an Euler pair, it follows that $2S_{1}\subseteq S_{1}$ and
$S_{2}=S_{1}\setminus 2S_{1}$. Given a partition $\lambda$ with parts in
$S_{1}$, denote by $\ell(\lambda)$ the number of parts in $\lambda$ and by
$\ell_{2}(\lambda)$ the number of even parts from $2S_{1}$.

Let $\mathcal{P}^{\prime}(S_{1},n)$ be the set of partitions $\lambda$ of $n$
with parts in $S_{1}$ such that $\lambda$ has at least one repeated part or at
least one part from $2S_{1}$ and denote by
$\mathcal{P}^{\prime}_{e}(S_{1},n)$, respectively
$\mathcal{P}^{\prime}_{o}(S_{1},n)$, the subset of partitions in
$\mathcal{P}^{\prime}(S_{1},n)$ with $\ell_{2}(\lambda)$ even, respectively
odd. We define an involution $\varphi$ on $\mathcal{P}^{\prime}(S_{1},n)$ that
reverses the parity of $\ell_{2}(\lambda)$.

Start with $\lambda\in\mathcal{P}^{\prime}(S_{1},n)$. Following Gupta’s
notation [10], we denote by $r$ the largest repeated part of $\lambda$ and by
$e$ the largest part of $\lambda$ that is from $2S_{1}$. If $r$ or $e$ do not
exist, we set them equal to $0$.

  1. (1)

If $2r>e$, we define $\varphi(\lambda)$ to be the partition obtained from
$\lambda$ by replacing two parts equal to $r$ by a single part equal to $2r$.

  2. (2)

If $2r\leqslant e$, we define $\varphi(\lambda)$ to be the partition obtained
from $\lambda$ by replacing one part equal to $e$ by two parts equal to $e/2$.

Since $\varphi:\mathcal{P}^{\prime}(S_{1},n)\to\mathcal{P}^{\prime}(S_{1},n)$
is an involution that reverses the parity of $\ell_{2}(\lambda)$, we have that
$|\mathcal{P}^{\prime}_{e}(S_{1},n)|=|\mathcal{P}^{\prime}_{o}(S_{1},n)|$.
This proves that

$q_{e}(S_{2};n)-q_{o}(S_{2};n)=p_{e}(S_{1},n)-p_{o}(S_{1},n).$

∎

######  Corollary 3.2.

Let $(S_{1},S_{2})$ be an Euler pair such that $S_{2}\subseteq 2\mathbb{N}-1$.
For $n\geqslant 0$, we have

$(-1)^{n}q(S_{2};n)=p_{e}(S_{1};n)-p_{o}(S_{1},n).$

###### Proof.

If $S_{2}$ consists of odd integers and $\lambda$ has parts in $S_{2}$, then
$\ell(\lambda)\equiv n\pmod{2}$. Then,
$q_{e}(S_{1};n)-q_{o}(S_{1};n)=(-1)^{n}q(S_{1};n)$. ∎

######  Remark 3.

It is possible to have an Euler pair $(S_{1},S_{2})$ such that
$S_{2}\not\subseteq 2\mathbb{N}-1$. For example, $S_{1}=\\{m\in\mathbb{N}\mod
m\equiv 2,4,5\pmod{6}\\}$ and $S_{2}=\\{m\in\mathbb{N}\mid m\equiv
2,5,11\pmod{12}\\}$ is an Euler pair and $q(S_{1};n)=p(S_{2};n)$ is Göllnitz’s
identity.

######  Remark 4.

If $S_{1}=\\{n\in\mathcal{N}\mid n\equiv\pm 1\pmod{3}\\}$ and
$S_{2}=\\{n\in\mathcal{N}\mid n\equiv\pm 1\pmod{6}\\}$, then $(S_{1},S_{2})$
is and Euler pair and $q(S_{1};n)=p(S_{2};n)$ is Schur’s identity. Thus, our
combinatorial proof of Theorem 3.1 gives a combinatorial proof of [15, Theorem
1.1]. Moreover, in this case $p(S_{1};n)=b_{3}(n)$ and $S_{2}\subseteq
2\mathbb{N}-1$. Thus, by Corollary 3.2, $b_{3}(n)\equiv q(S_{2};n)\pmod{2}$.

######  Theorem 3.3.

Let $r$ be an odd integer. For $n\geqslant 0$,

$q_{r}(\mathbb{N};n)\equiv q(M-r\,M;n)\pmod{2},$

where $M=2\mathbb{N}-1$,

###### Proof.

From Subbarao’s theorem (also Glaisher’s identity), we can write

$\displaystyle\sum_{n=0}^{\infty}q_{r}(\mathbb{N};n)\,q^{n}=\sum_{n=0}^{\infty}p(\mathbb{N}-r\,\mathbb{N};n)\,q^{n}=\frac{(q^{r};q^{r})_{\infty}}{(q;q)_{\infty}}.$

On the other hand, elementary techniques in the theory of partitions give the
following generating function

$\displaystyle F(z;q)$
$\displaystyle:=\prod_{k=0}^{\infty}\frac{1}{(1-zq^{kr+1})(1-zq^{kr+2})\cdots(1-zq^{kr+r-1})}$
$\displaystyle=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}p(\mathbb{N}-r\,\mathbb{N};n,m)\,z^{m}\,q^{n},$

where $p(\mathbb{N}-r\,\mathbb{N};n,m)$ is the number of partitions of $n$
with $m$ parts all taken from $\mathbb{N}-r\,\mathbb{N}$. We can write

$\displaystyle\sum_{n=0}^{\infty}(p_{e}(\mathbb{N}-r\,\mathbb{N};n)-p_{o}(\mathbb{N}$
$\displaystyle-r\,\mathbb{N};n))\,q^{n}=F(-1;q).$

On the other hand,

$\displaystyle F(-1;q)$
$\displaystyle=\frac{(-q^{r};q^{r})_{\infty}}{(-q;q)_{\infty}}=\frac{(q;q^{2})_{\infty}}{(q^{r};q^{2r})_{\infty}}$
$\displaystyle=\frac{(q;q^{2r})_{\infty}(q^{3};q^{2r})_{\infty}\cdots(q^{r};q^{2r})_{\infty}\cdots(q^{2r-1};q^{2r})_{\infty}}{(q^{r};q^{2r})_{\infty}}$
$\displaystyle=\sum_{n=0}^{\infty}(-1)^{n}\,q(M-rM;n)\,q^{n}.$

Thus we deduce that
$p_{e}(\mathbb{N}-r\,\mathbb{N};n)-p_{o}(\mathbb{N}-r\,\mathbb{N};n)=(-1)^{n}\,q(M-rM;n)$
and the proof follows easily. ∎

An overpartition of $n$ is a partition of $n$ in which the first occurrence of
a part may be overlined. If $(S_{1},S_{2})$ is an Euler pair of order $r$, we
denote by $\overline{p}_{r}(S_{1},n)$ the number of overpartitions of $n$ with
parts in $S_{1}$ and only parts from $S_{1}\setminus S_{2}$ may be overlined.

######  Theorem 3.4.

If $(S_{1},S_{2})$ is an Euler pair of order $r$, then

$q_{r}(S_{1};n)\equiv\overline{p}_{r}(S_{1},n)\pmod{2},$

###### Proof.

Since $(S_{1},S_{2})$ is an Euler pair of order $r$, we have
$q_{r}(S_{1};n)=p(S_{2};n)$. Moreover, $rS_{1}\subseteq S_{1}$ and
$S_{2}=S_{1}\setminus rS_{1}$. Thus

$q_{r}(S_{1};n)=p(S_{2};n)=p_{e}(S_{2};n)+p_{o}(S_{2};n)\equiv
p_{e}(S_{2};n)-p_{o}(S_{2};n)\pmod{2}.$

We have

$\displaystyle\sum_{n=0}^{\infty}(p_{e}(S_{2};n)-p_{o}(S_{2};n))q^{n}$
$\displaystyle=\prod_{i\in S_{2}}\frac{1}{1+q^{i}}=\prod_{i\in S_{1}\setminus
rS_{1}}\frac{1}{1+q^{i}}=\prod_{i\in S_{1}}\frac{1+q^{ri}}{1+q^{i}}$
$\displaystyle\equiv\prod_{i\in rS_{1}}(1+q^{i})\prod_{i\in
S_{1}}\frac{1}{1-q^{i}}\pmod{2}$

The first product generates the overlined parts and the second product
generates the non-overlined parts. Since $S_{2}=S_{1}\setminus rS_{1}$, this
concludes the proof. ∎

##  4\. Concluding remarks

In this article we found infinitely many new arithmetic progressions
$\\{s_{p,n}=p^{2}n+p\alpha-24_{p}^{-1}\\}_{n\geqslant 0}$, where
$p\in\mathcal{P}$ and $\alpha\neq\lfloor p/24\rfloor$ is a residue modulo $p$.
For small $p\in\mathcal{P}$, the congruneces could be proved using modular
forms. However, the general case required an intricate analysis of the
Diophantine equation

(9) $x^{2}+216y^{2}=pm,$

where $p\in\mathcal{P}$, $pm\equiv 1\pmod{24}$ and $p\nmid m$. The difficulty
in this analysis is caused by the fact that $216$ is not an idoneal or Euler
convenient number. To prove Theorem 1.3, we proved that the number of
primitive solutions for (9) is divisible by $8$.

Here, for $m\equiv 1\pmod{24}$, we conjecture an exact formula for $N_{2}(m)$,
the number of primitive solutions for

(10) $x^{2}+216y^{2}=m.$

From [5, Lemma 1.7], if there is a prime $p\mid m$ with $p\equiv
13,17,19,23\pmod{24}$, then $N_{2}(m)=0$.

Denote by $\mathcal{Q}$ be the set of prime numbers not in $\mathcal{P}$.
Given a positive integer $m$, 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 $p_{1},p_{2},\ldots,p_{k}\in\mathcal{P}$ and
$q_{1},q_{2},\ldots,q_{n}\in\mathcal{Q}$. Let

$B(m):=|\\{b_{i}\in\mathcal{Q}:b_{i}\equiv 0\pmod{3}\\}|.$

######  Conjecture 1.

Let $m$ be a positive integer such that $m\equiv 1\pmod{24}$ and all prime
divisors of $m$ are congruent to $1,5,7,11$ modulo $24$. With the above
notation we have

$\begin{split}N_{2}(m)=&2^{B(m)+k}\cdot\frac{2^{n-B(m)+1}+4\cdot(-1)^{n-B(m)}}{3}\\\
=&\frac{2^{n+k+1}+2^{B(m)+k+2}\cdot(-1)^{n-B(m)}}{3}.\end{split}$

We note that Conjecture 1 implies Theorem 1.1.

As noted in Remark 4, it follows from Corollary 3.2 that

$b_{3}(n)\equiv q(S_{2};n)\pmod{2},$

where $q(S_{2};n)$ is the number of distinct $3$-regular partitions with all
parts odd. Then the congruences of Theorems 1.1, 1.2, and 1.3 hold for
$q(S_{2};n)$.

The results of section 3 led us to search for parity results for $b_{3,e}(n)$
(respectively $b_{3,e}(n)$), the number of $3$-regular partitions with an even
(respectively odd) number of parts.

We end the article with the following congruence conjectures for which there
is substantial numerical evidence.

The first conjecture is similar to Theorem 1.1.

######  Conjecture 2.

For $n\geqslant 0$ and $p\equiv 13,17,19,23\pmod{24}$,

$b_{3,e}\big{(}2\,(p^{2}\,n+p\alpha-24^{-1})\big{)}\equiv 0\pmod{2},$
$b_{3,o}\big{(}2\,(p^{2}\,n+p\alpha-24^{-1})\big{)}\equiv 0\pmod{2},$

where $1\leqslant\alpha\leqslant p-1$ and $24^{-1}$ is taken modulo $p^{2}$.

The next conjecture is similar to Theorem 1.3.

######  Conjecture 3.

For $n\geqslant 0$, and $p\in\mathcal{P}$ with $p\equiv 1,7\pmod{24}$,

$b_{3,e}\big{(}2\,(p^{2}\,n+p\,\alpha-24_{p}^{-1})\big{)}\equiv 0\pmod{2},$
$b_{3,o}\big{(}2\,(p^{2}\,n+p\,\alpha-24_{p}^{-1})\big{)}\equiv 0\pmod{2},$

where $0\leqslant\alpha<p$, $\alpha\neq\lfloor p/24\rfloor$.

To prove Conjectures 2 and 3, one could try to make use of Corollary 3.2 which
gives

$2b_{3,e}(m)=b_{3}(m)+q(S_{2};m).$

Thus, to prove the conjectures, it is enough to show that

$b_{3}(m)\equiv q(S_{2};m)\pmod{4}$

for the suitable values of $m$.

A different approach would be to use [15, Theorem 1.4], which states that

$\displaystyle b_{3,e}(m)$ $\displaystyle=p(S_{3};m),$

where

$\displaystyle S_{3}$ $\displaystyle=\\{n\in\mathbb{N}\mid n\not\equiv 0,\pm
1,\pm 10,\pm 11,12\mod{24}\\},$

and try to prove the conjectures for the function $p(S_{3};m)$.

It is likely that similar conjectures can be formulated as companions to
Theorem 1.1. For example, there is considerable numerical evidence for the
next conjecture which fits into the case $k=0$ of Theorem 1.1 (2).

######  Conjecture 4.

For $n\geqslant 0$, and $p\in\\{7,31\\}$,

$b_{3,e}\big{(}2\,(p^{3}\,n+p^{2}\alpha-24^{-1})\big{)}\equiv 0\pmod{2};$
$b_{3,o}\big{(}2\,(p^{3}\,n+p^{2}\alpha-24^{-1})\big{)}\equiv 0\pmod{2};$

where $0\leqslant\alpha<p$, $\alpha\not\equiv-24^{-1}\pmod{p}$, and $24^{-1}$
is taken modulo $p^{2}$.

## References

  * [1] G. E. Andrews, Two theorems of Euler and a general partition theorem, Proc. Amer. Math. Soc. 20 (1969), 499–502.
  * [2] G. E. Andrews, The Theory of Partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998. Reprint of the 1976 original.
  * [3] R. Carlson, J. J. Webb, Infinite families of infinite families of congruences for $k$-regular partitions, Ramanujan J 33 (2014) 329–337.
  * [4] S. Chern, New congruences for $2$-color partitions. J. Number Theory 163 (2016), 474–481.
  * [5] D. Cox, Primes of the form $x^{2}+ny^{2}$. Fermat, class field theory, and complex multiplication, Second edition. Pure and Appl. Math., John Wiley & Sons, Inc., Hoboken, NJ, 2013.
  * [6] S.-P. Cui, N. S. S. Gu, Arithmetic properties of $\ell$-regular partitions, Adv. Appl. Math. 51(4) (2013) 507–523.
  * [7] B. Dandurand, D. Penninston, $\ell$-Divisibility of $\ell$-regular partition functions, Ramanujan J 19 (2009) 63–70.
  * [8] L. Dickson, Introduction to the Theory of Numbers, University of Chicago Press, Chicago, 1929.
  * [9] D. Furcy, D. Penniston, Congruences for $\ell$-regular partition functions modulo $3$, Ramanujan J 27 (2012) 101–108.
  * [10] H. Gupta, Combinatorial proof of a theorem on partitions into an even or odd number of parts. J. Combinatorial Theory Ser. A 21 (1976), no. 1, 100–103.
  * [11] M. D. Hirschhorn, J. A. Sellers, Elementary proofs of parity results for $5$-regular partitions, Bull. Aust. Math. Soc. 81(1) (2010) 58–63.
  * [12] G. James, A. Kerber, The Representation of the Symmetric Group, Addison-Wesley, Reading, 1979.
  * [13] W. J. Keith and F. Zanello, Parity of the coefficients of certain eta-quotients, J. Number Theory 235 (2022), 275– 304
  * [14] J. Lovejoy, D. Penniston, $3$-regular partitions and a modular $K3$ surface, Contemp. Math. 291 (2001) 177–182.
  * [15] M. Merca, A reversal of Schur’s partition theorem. Rev. Real Acad. Cienc. Exactas Fis. Nat. Ser. A-Mat. 116(4) (2022) Paper No. 181.
  * [16] D. Penniston, The $p^{a}$-regular partition function modulo $p^{j}$, J. Number Theory 94 (2002) 320–325.
  * [17] D. Penniston, Arithmetic of $\ell$-regular partition functions, Int. J. Number Theory 4 (2008) 295–302.
  * [18] S. Radu, An algorithmic approach to Ramanujan’s congruences, Ramanujan J. 20(2) (2009) 215–251.
  * [19] C. S. Radu and J. A. Sellers, Congruence properties modulo 5 and 7 for the pod function. Int. J. Number Theory 7 (2011), no. 8, 2249–2259.
  * [20] M. V. Subbarao, Partition theorems for Euler pairs. Proc. Amer. Math. Soc. 28 (1971), 330–336.
  * [21] L. Wang, Arithmetic properties of $(k,\ell)$-regular bipartitions. Bull. Aust. Math. Soc. 95 (2017), no. 3, 353–364.
  * [22] J. J. Webb, Arithmetic of the $13$-regular partition function modulo $3$, Ramanujan J 25 (2011) 49–56.
  * [23] E. X. W. Xia, Congruences for some $\ell$-regular partitions modulo $l$, J. Number Theory 152 (2015) 105–117.
  * [24] E. X. W. Xia and O. X. M. Yao, Parity results for $9$-regular partitions, Ramanujan J 34 (2014) 109–117.
  * [25] O. X. M. Yao, New parity results for $3$-regular partitions, Quaestiones Mathematicae, (2022) DOI: 10.2989/16073606.2022.2033872