Odd 4-Perfect Numbers Exactly Divisible by $3^2$

Abstract

We call a positive integer N a 4-perfect number if the sum of all divisors of N is equal to 4N. In this paper, we give some sufficient conditions such that an odd integer exactly divisible by $3^2$ cannot be 4-perfect. Moreover, we refine Euler’s structure theorem on odd 4-perfect numbers given by Broughan and Zhou in the case where N is exactly divisible by $3^2$.

1 Introduction

For a given positive integer $k \ge 2$, we call a positive integer N a multiperfect number of abundancyk or k-perfect number if the sum $\sigma (N)$ of all divisors of N is equal to kN. By the definition, a 2-perfect number is a perfect number. Descartes, Fermat, and Jumeau had already found several multiperfect numbers in the seventeenth century. As of 2017, more than 5000 multiperfect numbers are known, which are all even. As well as odd perfect numbers, the existence of odd multiperfect numbers is an unsolved problem. Many people have studied necessary conditions for the existence of odd multiperfect numbers, for example [1, 3,4,5,6,7,8,9]. In 2008, Broughan and Zhou gave Euler’s structure theorem on odd 4-perfect numbers.

Theorem 1

([2, Theorem 2.1]) Let N be an odd 4-perfect number. Then, N has one of the following forms, where $\alpha _1, \alpha _2, \ldots , \alpha _r$ are even positive integers and $p_1, p_2, \ldots , p_r$ are odd prime numbers.

• (A) $N =q_1^{e_1} q_2^{e_2} p_1^{\alpha _1} p_2^{\alpha _2} \cdots p_r^{\alpha _r}$ for prime numbers $q_j$ and positive integers $e_j$ with $q_j \equiv e_j \equiv 1\ \text{(mod } \ 4\text{) }$. In the remaining types, $N =q^{e} p_1^{\alpha _1} p_2^{\alpha _2} \cdots p_r^{\alpha _r}$ where

• (B) $q \equiv 1\ \text{(mod } \ 4\text{) }$ and $e \equiv 3\ \text{(mod } \ 8\text{) }$, or

• (C) $q \equiv 3\ \text{(mod } \ 8\text{) }$ and $e \equiv 1\ \text{(mod } \ 4\text{) }$.

Broughan and Zhou showed also the following theorem.

Theorem 2

([2, Theorem 2.4]) If N is a positive integer exactly divisible by $3^2$ and such that if 13,61, and 97 appear in the prime factorization of N, then their powers are congruent to 2 modulo 6. Then, N is not an odd 4-perfect number.

Our main purpose in this paper is to refine Theorem 1 in the case where N is exactly divisible by $3^2$ (Sect. 3, Theorems 5, 6, and 7). For the refinement, we look into the order of the divisor function at 3 (Sect. 2, Lemma 2). Theorem 2 follows from Lemma 2 immediately. Practically, we give a generalization of Theorem 2.

Theorem 3

If N is a positive integer exactly divisible by $3^2$ and satisfies either of the following conditions, then N is not an odd 4-perfect number.

• (1) At least one prime factor of N is congruent to 1 modulo 3, and its power is congruent to 26 modulo 27.

• (2) At least two prime factors of N are congruent to 1 modulo 3, and one of their powers is congruent to 2 modulo 3 and the other is congruent to 8 modulo 9.

• (3) At least three prime factors of N are congruent to 1 modulo 3, and their powers are congruent to 2 modulo 3.

We consider also the case where some prime factors of N are congruent to 2 modulo 3 and their powers are odd. Note that an odd 4-perfect number has at most two prime factors with odd exponent by Theorem 1. We show that Lemma 2 implies the following theorem.

Theorem 4

If N is a positive integer exactly divisible by $3^2$ and satisfies either of the following conditions, then N is not an odd 4-perfect number.

• (1) At least one prime factor of N is congruent to 2 modulo 3, and its power is congruent to 17 modulo 18.

• (2) At least two prime factors of N are congruent to 2 modulo 3, and one of their powers is odd and the other is congruent to 5 modulo 6.

2 The Order of the Divisor Function at 3

For a positive integer N and a prime number p, if N is divisible by $p^e$ and not divisible by $p^{e+1}$, then we denote the nonnegative integer e by $\mathrm {ord}_p(N)$. We use the following lemmas.

Lemma 1

(e.g., [2, Lemma 2.2]) Let $p \ge 5$ be a prime number, and let e be a positive integer. If $p \equiv 1\ \text{(mod } \ 3\text{) }$, then

$$\begin{aligned} \sigma (p^e) \equiv {\left\{ \begin{array}{ll} 1 \pmod {3}, &{}\quad e \equiv 0 \pmod {3}; \\ 2 \pmod {3}, &{}\quad e \equiv 1 \pmod {3}; \\ 0 \pmod {3}, &{}\quad e \equiv 2 \pmod {3}. \end{array}\right. } \end{aligned}$$

If $p \equiv 2\ \text{(mod } \ 3\text{) }$, then

$$\begin{aligned} \sigma (p^e) \equiv {\left\{ \begin{array}{ll} 1 \pmod {3}, &{}\quad e \equiv 0 \pmod {2}; \\ 0 \pmod {3}, &{}\quad e \equiv 1 \pmod {2}. \end{array}\right. } \end{aligned}$$

Lemma 2

Let p be an odd prime number, and let e be a positive integer. If $p \equiv 1 \pmod {3}$, then $\mathrm {ord}_3(\sigma {(p^e)})=\mathrm {ord}_3(e+1)$. If $p \equiv 2 \pmod {3}$, then

$$\begin{aligned} \mathrm {ord}_3(\sigma {(p^e)})= {\left\{ \begin{array}{ll} \mathrm {ord}_3(p+1)+\mathrm {ord}_3(e+1), &{}\quad \text {if } e \text { is odd};\\ 0, &{}\quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Let u and k be integers such that $e+1=3^uk$ with $\mathrm {gcd}(k, 3)=1$. Then, we have

$$\begin{aligned} \sigma {(p^e)}&=1+p+p^2+ \cdots + p^{e}\\&=(1+p+p^2+ \cdots + p^{k-1}) \prod _{l=0}^{u-1}(1+p^{3^lk}+p^{2\cdot 3^lk}). \end{aligned}$$

If $p \equiv 1 \pmod {3}$, then $1+p+p^2+ \cdots + p^{k-1} \equiv k \not \equiv 0\ \text{(mod } \ 3\text{) }$. Let $A_l$ be an integer such that $p^{3^lk}=3A_l+1$. Then, we have

$$\begin{aligned} 1+p^{3^lk}+p^{2\cdot 3^lk}&= 1+(3A_l+1)+(3A_l+1)^2 \\&= 3(3A_l^2+3A_l+1) \\&\equiv 3\pmod {9}. \end{aligned}$$

Therefore, $\mathrm {ord}_3(1+p+p^2+ \cdots + p^{k-1})=0$ and $\mathrm {ord}_3(1+p^{3^lk}+p^{2\cdot 3^lk}) = 1$ for each l. We see that $\mathrm {ord}_3(\sigma {(p^e)})=u=\mathrm {ord}_3(e+1)$.

If $p \equiv 2 \pmod {3}$ and e is even, then we have $\sigma {(p^e)} \equiv 1\ \text{(mod } \ 3\text{) }$ by Lemma 1. Therefore, $\mathrm {ord}_3(\sigma {(p^e)})=0$. Next, we assume that $p \equiv 2 \pmod {3}$ and e is odd. Then, k is even. Moreover, k is not divisible by 3 by our assumption. Thus, we have

$$\begin{aligned} 1+p+p^2+ \cdots + p^{k-1} = (1+p)(1+p^2+p^4+ \cdots + p^{k-2}) \end{aligned}$$

and

$$\begin{aligned} 1+p^2+p^4+ \cdots + p^{k-2} \equiv \frac{k}{2} \not \equiv 0 \pmod {3}. \end{aligned}$$

Since k is even, we have $p^{3^lk} \equiv 1\ \text{(mod } \ 3\text{) }$. Let $B_l$ be an integer such that $p^{3^lk}=3B_l+1$. Then, we have

$$\begin{aligned} 1+p^{3^lk}+p^{2\cdot 3^lk}&= 1+(3B_l+1)+(3B_l+1)^2 \\&= 3(3B_l^2+3B_l+1) \\&\equiv 3 \pmod {9}. \end{aligned}$$

Therefore, $\mathrm {ord}_3(1+p+p^2+ \cdots + p^{k-1})=\mathrm {ord}_3(1+p)$ and $\mathrm {ord}_3(1+p^{3^lk}+p^{2\cdot 3^lk}) = 1$ for each l. We see that $\mathrm {ord}_3(\sigma {(p^e)})=\mathrm {ord}_3(p+1)+\mathrm {ord}_3(e+1)$. $\square$

We show that Lemma 2 implies Theorems 3 and 4 in Introduction. Let N be a positive integer exactly divisible by $3^2$. By Lemma 2, we see that $\mathrm {ord}_3(\sigma (N)) \ge 3$ under our assumptions. Note that $\sigma (ab)=\sigma (a)\sigma (b)$ if a is coprime to b. On the other hand, $\mathrm {ord}_3(4N) =2$. Therefore, N cannot be an odd 4-perfect number.

3 The Structure Theorems

We give the structure theorems on odd 4-perfect numbers exactly divisible by $3^2$. We classify into 3 theorems according to a prime factor form of N given in Theorem 1.

Theorem 5

We assume that an odd 4-perfect number N is exactly divisible by $3^2$ and has the form of $\mathrm {(A)}$ in Theorem 1. Then, N has one of the following forms, where $r \ge 9$.

• (A1) $N =q_1^{e_1} q_2^{e_2} 3^2p_2^{\alpha _2}\cdots p_r^{\alpha _r}$ for prime numbers $q_j, p_i$ and positive integers $e_j, \alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q_j\equiv 1\ \text{(mod } \ 12\text{) } \hbox { and } e_j\equiv 17, 89\ \text{(mod } \ 108\text{) };\\ q_j\equiv 5, 29\ \text{(mod } \ 36\text{) } \hbox { and } e_j\equiv 5, 29\ \text{(mod } \ 36\text{) };\\ q_j\equiv 17, 89\ \text{(mod } \ 108\text{) } \hbox { and } e_j\equiv 1, 9\ \text{(mod } \ 12\text{) },\\ \end{array}\right. } \\&\;\;\; q_k\equiv 1\ \text{(mod } \ 12\text{) } \hbox { and } e_k \equiv 1, 9\ \text{(mod } \ 12\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\hbox { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\hbox { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) }, \end{array}\right. } \end{aligned}$$

  where $(j,k)=(1,2)$ or (2, 1), and $i=2, \ldots , r$.

• (A2) $N =q_1^{e_1} q_2^{e_2} 3^2p_2^{\alpha _2}\cdots p_r^{\alpha _r}$ for prime numbers $q_j, p_i$ and positive integers $e_j, \alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q_j\equiv 1\ \text{(mod } \ 12\text{) } \text { and } e_j \equiv 5, 29\ \text{(mod } \ 36\text{) };\\ q_j\equiv 5, 29\ \text{(mod } \ 36\text{) } \text { and } e_j \equiv 1, 9\ \text{(mod } \ 12\text{) }, \\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) } \text { and } \alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) } \text { and } \alpha _i \equiv 0\ \text{(mod } \ 2\text{) }, \end{array}\right. } \end{aligned}$$

  where $j=1,2$ and $i=2, \ldots , r$.

• (A3) $N =q_1^{e_1} q_2^{e_2} 3^2 p^f p_3^{\alpha _3}\cdots p_r^{\alpha _r}$ for prime numbers $q_j, p, p_i$ and positive integers $e_j, f,\alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q_j\equiv 1\ \text{(mod } \ 12\text{) } \text { and } e_j\equiv 5, 29\ \text{(mod } \ 36\text{) };\\ q_j\equiv 5, 29\ \text{(mod } \ 36\text{) } \text { and } e_j\equiv 1, 9\ \text{(mod } \ 12\text{) }, \end{array}\right. }\\&\;\;\; q_k\equiv 1\ \text{(mod } \ 12\text{) } \text { and } e_k \equiv 1, 9\ \text{(mod } \ 12\text{) },\\&\;\;\; p \equiv 1\ \text{(mod } \ 3\text{) } \text { and } f \equiv 2, 14\ \text{(mod } \ 18\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $(j,k)=(1,2)$ or (2, 1), and $i=3, \ldots , r$.

• (A4) $N =q_1^{e_1} q_2^{e_2} 3^2p^f p_3^{\alpha _3}\cdots p_r^{\alpha _r}$ for prime numbers $q_j, p, p_i$ and positive integers $e_j, f, \alpha _i$ with

  $$\begin{aligned}&\;\;\; q_j \equiv 1\ \text{(mod } \ 12\text{) }\text { and }e_j \equiv 1, 9\ \text{(mod } \ 12\text{) },\\&\;\;\; p \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f \equiv 8, 44\ \text{(mod } \ 54\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) }, \end{array}\right. } \end{aligned}$$

  where $j=1,2$ and $i=3, \ldots , r$.

• (A5) $N =q_1^{e_1} q_2^{e_2}3^2 l_1^{f_1} l_2^{f_2}p_4^{\alpha _4} \cdots p_r^{\alpha _r}$ for prime numbers $q_j, l_m, p_i$ and positive integers $e_j, f_m, \alpha _i$ with

  $$\begin{aligned}&\;\;\; q_j \equiv 1\ \text{(mod } \ 12\text{) }\text { and }e_j \equiv 1, 9\ \text{(mod } \ 12\text{) },\\&\;\;\; l_m \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f_m \equiv 2, 14\ \text{(mod } \ 18\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) }, \end{array}\right. } \end{aligned}$$

  where $j=1,2$, $m=1,2$, and $i=4, \ldots , r$.

Proof

If N is exactly divisible by $3^2$, then $\mathrm {ord}_3(N)=2$. Moreover, $\mathrm {ord}_3(\sigma (N)) = 2$ because $\sigma (N)=4N$. Let $N= q_1^{e_1} q_2^{e_2}p_1^{\alpha _1} p_2^{\alpha _2}\cdots p_r^{\alpha _r}$ be a prime factorization of N, where $p_1=3$, $\alpha _1=2$, $q_j \equiv e_j \equiv 1 \pmod {4}$ for $j=1,2$, and $r \ge 9$ by [4]. Since $\mathrm {ord}_3(\sigma (3^2)) =0$, we have

$$\begin{aligned} 2&=\mathrm {ord}_3(\sigma (N))\\&=\mathrm {ord}_3\left( \sigma \left( q_1^{e_1} q_2^{e_2}3^2p_2^{\alpha _2}\cdots p_r^{\alpha _r}\right) \right) \\&=\mathrm {ord}_3\left( \sigma \left( q_1^{e_1}\right) \right) +\mathrm {ord}_3\left( \sigma \left( q_2^{e_2}\right) \right) +\mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) +\cdots +\mathrm {ord}_3\left( \sigma \left( p_r^{\alpha _r}\right) \right) . \end{aligned}$$

We need to consider the following five cases.

(A1)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q_j^{e_j}\right) \right) =2;\\ \mathrm {ord}_3\left( \sigma \left( q_k^{e_k}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $(j,k)=(1,2)$ or (2, 1), and $i=2, \ldots , r$.

(A2)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q_1^{e_1}\right) \right) =\mathrm {ord}_3\left( \sigma \left( q_2^{e_2}\right) \right) =1;\\ \mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $i=2, \ldots , r$.

(A3)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q_j^{e_j}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =1;\\ \mathrm {ord}_3\left( \sigma \left( q_k^{e_k}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $(j,k)=(1,2)$ or (2, 1), and $i=3, \ldots , r$.

(A4)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q_1^{e_1}\right) \right) =\mathrm {ord}_3\left( \sigma \left( q_2^{e_2}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0;\\ \mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =2, \end{array}\right. }$

where $i=3, \ldots , r$.

(A5)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q_1^{e_1}\right) \right) =\mathrm {ord}_3\left( \sigma \left( q_2^{e_2}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0;\\ \mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_3^{\alpha _3}\right) \right) =1, \end{array}\right. }$

where $i=4, \ldots , r$.

By combining $\mathrm {(A1)'}$ with Lemma 2 (1), (3), and Theorem 1$\mathrm {(A)}$, we obtain the condition $\mathrm {(A1)}$. Similarly, we obtain each condition $\mathrm {(A2)}$, $\mathrm {(A3)}$, $\mathrm {(A4)}$, and $\mathrm {(A5)}$ by combining $\mathrm {(A2)'}$, $\mathrm {(A3)'}$, $\mathrm {(A4)'}$, and $\mathrm {(A5)'}$ with Lemma 2 and Theorem 1 (A), respectively. $\square$

Theorem 6

We assume that an odd 4-perfect number N is exactly divisible by $3^2$ and has the form of $\mathrm {(B)}$ in Theorem 1. Then, N has one of the following forms, where $r \ge 9$.

• (B1) $N= q^e 3^2 p_2^{\alpha _2} \cdots p_r^{\alpha _r}$ for prime numbers $q, p_i$ and positive integers $e, \alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q\equiv 1\ \text{(mod } \ 12\text{) }\text { and }e\equiv 35, 179\ \text{(mod } \ 216\text{) };\\ q\equiv 5, 29\ \text{(mod } \ 36\text{) }\text { and }e\equiv 11, 59\ \text{(mod } \ 72\text{) };\\ q\equiv 17, 89\ \text{(mod } \ 108\text{) }\text { and }e\equiv 3, 19\ \text{(mod } \ 24\text{) }, \end{array}\right. }\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) }, \\ \end{array}\right. } \end{aligned}$$

  where $i=2, \ldots , r$.

• (B2) $N= q^e 3^2 p^f p_3^{\alpha _3} \cdots p_r^{\alpha _r}$ for prime numbers $q, p, p_i$ and positive integers $e, f, \alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q\equiv 1\ \text{(mod } \ 12\text{) }\text { and }e\equiv 11, 59\ \text{(mod } \ 72\text{) };\\ q\equiv 5, 29\ \text{(mod } \ 36\text{) }\text { and }e\equiv 3, 19\ \text{(mod } \ 24\text{) },\\ \end{array}\right. }\\&\;\;\; p \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f \equiv 2, 14\ \text{(mod } \ 18\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $i=3, \ldots , r$.

• (B3) $N= q^e 3^2 p^f p_3^{\alpha _3} \cdots p_r^{\alpha _r}$ for prime numbers $q, p, p_i$ and positive integers $e, f, \alpha _i$ with

  $$\begin{aligned}&\;\;\; q\equiv 1\ \text{(mod } \ 12\text{) }\text { and }e\equiv 3, 19\ \text{(mod } \ 24\text{) };\\&\;\;\; p \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f \equiv 8, 44\ \text{(mod } \ 54\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $i=3, \ldots , r$.

• (B4) $N= q^e 3^2 l_1^{f_1} l_2^{f_2} p_4^{\alpha _4} \cdots p_r^{\alpha _r}$ for prime numbers $q, l_j, p_i$ and positive integers $e, f_j, \alpha _i$ with

  $$\begin{aligned}&\;\;\; q\equiv 1\ \text{(mod } \ 12\text{) }\text { and }e\equiv 3, 19\ \text{(mod } \ 24\text{) };\\&\;\;\; l_j \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f_j \equiv 2, 14\ \text{(mod } \ 18\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $j=1,2$ and $i=4, \ldots , r$.

Proof

Let $N= q^ep_1^{\alpha _1} p_2^{\alpha _2}\cdots p_r^{\alpha _r}$ be a prime factorization of N, where $p_1=3$, $\alpha _1=2$, $r \ge 9$, $q \equiv 1 \pmod {4}$, $e \equiv 3 \pmod {8}$. In a similar way to the proof of Theorem 5, we have

$$\begin{aligned} 2 =\mathrm {ord}_3\left( \sigma \left( q^e\right) \right) +\mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) +\cdots +\mathrm {ord}_3\left( \sigma \left( p_r^{\alpha _r}\right) \right) . \end{aligned}$$

We need to consider the following four cases.

(B1)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q^e\right) \right) =2;\\ \mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $i=2, \ldots , r$.

(B2)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q^e\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_2^{e_2}\right) \right) =1;\\ \mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $i=3, \ldots , r$.

(B3)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q^e\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0;\\ \mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =2, \end{array}\right. }$

where $i=3, \ldots , r$.

(B4)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3\left( \sigma \left( q^e\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0;\\ \mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_3^{\alpha _3}\right) \right) =1, \end{array}\right. }$

where $i=4, \ldots , r$.

We obtain each condition $\mathrm {(B1)}$, $\mathrm {(B2)}$, $\mathrm {(B3)}$, and $\mathrm {(B4)}$ by combining $\mathrm {(B1)'}$, $\mathrm {(B2)'}$, $\mathrm {(B3)'}$, and $\mathrm {(B4)'}$ with Lemma 2 and Theorem 1$\mathrm {(B)}$ respectively. $\square$

Theorem 7

We assume that an odd 4-perfect number N is exactly divisible by $3^2$ and has the form of $\mathrm {(C)}$ in Theorem 1. Then, N has one of the following forms, where $q, p_1, \ldots p_r, p, l_1, l_2$ are odd prime numbers.

• (C1) $N=q^e3^2p_2^{\alpha _2} \cdots p_r^{\alpha _r}$ for prime numbers $q, p_i$ and positive integers $e, \alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q\equiv 19\ \text{(mod } \ 24\text{) }\text { and }e \equiv 17, 89\ \text{(mod } \ 108\text{) };\\ q \equiv 11, 59\ \text{(mod } \ 72\text{) }\text { and }e\equiv 5, 29\ \text{(mod } \ 36\text{) };\\ q\equiv 35, 179\ \text{(mod } \ 216\text{) }\text { and }e\equiv 1, 9\ \text{(mod } \ 12\text{) }, \end{array}\right. }\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) },\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $i=2, \ldots , r$.

• (C2) $N=q^e3^2p^fp_3^{\alpha _3} \cdots p_r^{\alpha _r}$ for prime numbers $q, p, p_i$ and positive integers $e, f, \alpha _i$ with

  $$\begin{aligned}&{\left\{ \begin{array}{ll} q\equiv 19\ \text{(mod } \ 24\text{) }\text { and }e\equiv 5, 29\ \text{(mod } \ 36\text{) };\\ q\equiv 11, 59\ \text{(mod } \ 72\text{) }\text { and }e\equiv 1, 9\ \text{(mod } \ 12\text{) }, \end{array}\right. }\\&\;\;\; p\equiv 1\ \text{(mod } \ 3\text{) }\text { and }f \equiv 2, 14\ \text{(mod } \ 18\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $i=3, \ldots , r$.

• (C3) $N=q^e3^2p^fp_3^{\alpha _3} \cdots p_r^{\alpha _r}$ for prime numbers $q, p, p_i$ and positive integers $e, f, \alpha _i$ with

  $$\begin{aligned}&\;\;\; q\equiv 19\ \text{(mod } \ 24\text{) }\text { and }e\equiv 1, 9\ \text{(mod } \ 12\text{) }, \\&\;\;\; p \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f \equiv 8, 44\ \text{(mod } \ 54\text{) },\\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) },\\ \end{array}\right. } \end{aligned}$$

  where $i=3, \ldots , r$.

• (C4) $N=q^e3^2l_1^{f_1}l_2^{f_2}p_4^{\alpha _4} \cdots p_r^{\alpha _r}$ for prime numbers $q, l_j, p_i$ and positive integers $e, f_j, \alpha _i$ with

  $$\begin{aligned}&\;\;\; q\equiv 19\ \text{(mod } \ 24\text{) }\text { and }e\equiv 1, 9\ \text{(mod } \ 12\text{) }, \\&\;\;\; l_j \equiv 1\ \text{(mod } \ 3\text{) }\text { and }f_j \equiv 2, 14\ \text{(mod } \ 18\text{) }, \\&{\left\{ \begin{array}{ll} p_i \equiv 1\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0, 4\ \text{(mod } \ 6\text{) };\\ p_i \equiv 2\ \text{(mod } \ 3\text{) }\text { and }\alpha _i \equiv 0\ \text{(mod } \ 2\text{) }, \end{array}\right. } \end{aligned}$$

  where $j=1,2$ and $i=4, \ldots , r$.

Proof

Let $N= q^ep_1^{\alpha _1} p_2^{\alpha _2}\cdots p_r^{\alpha _r}$ be a prime factorization of N, where $p_1=3$, $\alpha _1=2$, $r \ge 9$, $q \equiv 3 \pmod {8}$, $e \equiv 1 \pmod {4}$. In a similar way to the proof of Theorem 5, we have

$$\begin{aligned} 2=\mathrm {ord}_3\left( \sigma \left( q^e\right) \right) +\mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) +\cdots +\mathrm {ord}_3\left( \sigma \left( p_r^{\alpha _r}\right) \right) . \end{aligned}$$

We need to consider the following four conditions.

(C1)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3(\sigma (q^e))=2;\\ \mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $i=2, \ldots , r$.

(C2)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3(\sigma (q^e))=\mathrm {ord}_3\left( \sigma \left( p_2^{e_2}\right) \right) =1; \\ \mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0, \end{array}\right. }$

where $i=3, \ldots , r$.

(C3)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3(\sigma (q^e))=\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0;\\ \mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =2, \end{array}\right. }$

where $i=3, \ldots , r.$

(C4)$'$:

${\left\{ \begin{array}{ll} \mathrm {ord}_3(\sigma (q^e))=\mathrm {ord}_3\left( \sigma \left( p_i^{\alpha _i}\right) \right) =0;\\ \mathrm {ord}_3\left( \sigma \left( p_2^{\alpha _2}\right) \right) =\mathrm {ord}_3\left( \sigma \left( p_3^{\alpha _3}\right) \right) =1, \end{array}\right. }$

where $i=4, \ldots , r$.

We obtain each condition $\mathrm {(C1)}$, $\mathrm {(C2)}$, $\mathrm {(C3)}$, and $\mathrm {(C4)}$ by combining $\mathrm {(C1)'}$, $\mathrm {(C2)'}$, $\mathrm {(C3)'}$, and $\mathrm {(C4)'}$ with Lemma 2 and Theorem 1$\mathrm {(C)}$ respectively. $\square$