On a problem of De Koninck

TL;DR

This paper investigates the solutions to the equation involving the sum of divisors and the product of prime divisors of an integer, identifying specific prime divisibility conditions for all solutions except two special cases.

Contribution

It characterizes the structure of integers satisfying (n) = ((n))^2, revealing prime divisibility patterns and excluding all but two known solutions.

Findings

Identifies all solutions n (n) = (
6(n))^2 except for n=1 and n=1782.

Describes the prime divisibility conditions that solutions must satisfy.

Provides a structural description of solutions involving chains of primes and divisor sums.

Abstract

Let $\sigma(n)$ and $\gamma(n)$ denote the sum of divisors and the product of distinct prime divisors of $n$ respectively. We shall show that, if $n\neq 1, 1782$ and $\sigma(n)=(\gamma(n))^2$, then there exist odd (not necessarily distinct) primes $p, p^\prime$ and (not necessarily odd) distinct primes $q_i (i=1, 2, \ldots, k)$ such that $p, p^\prime\mid\mid n$, $q_i^2\mid\mid n (i=1, 2, \ldots, k)$ and $q_1\mid \sigma(p^2), q_{i+1}\mid\sigma(q_i^2) (1\leq i\leq k-1), p^\prime \mid\sigma(q_k^2)$.

Equations

$$\sigma (n)=(\gamma (n){)}^2$$

$$n=2^{e_0}\prod _{i=1}^s{p}_i^{e_i},$$

$$n<\exp \left(((K+\log L)T!{)}^{2^T}\right).$$

$$\sum _{\begin{array}{c}P:v_k\to v_{k-1}\to \cdots \to v_0\subset G,\\ v_k\in S(G)\end{array}}\frac{1}{{\prod }_{i=0}^{k-1}{d}_G^{-}(v_i)}=1.$$

$$\sum _{\begin{array}{c}{v}_k\in S(G),\\ v_k\to v_{k-1}\to \cdots \to v_0\subset G\end{array}}\frac{1}{{\prod }_{i=0}^{k-1}{d}_G^{-}(v_i)}=\frac{1}{{\prod }_{i=0}^{k-2}{d}_G^{-}(v_i)}.$$

$$\sum _{\begin{array}{c}P:v_k\to v_{k-1}\to \cdots \to v_0\subset G,\\ v_k\in S(G)\end{array}}\frac{1}{{\prod }_{i=0}^{k-1}{d}_G^{-}(v_i)}=\sum _{\begin{array}{c}P:v_k\to v_{k-1}\to \cdots \to v_0\subset H,\\ v_k\in S(H)\end{array}}\frac{1}{{\prod }_{i=0}^{k-1}{d}_H^{-}(v_i)}.$$

$$\sigma (p_i^{e_i})={\kappa }_i\prod _{p_j\in N(L)}{p}_j^{k_{i,j}}$$

$$p_i^2={\lambda }_i\prod _{p_j\in N(L)}{p}_j^{k_{i,j}},$$

$$\prod _{p_i\in B}\sigma (p_i^{e_i})=\prod _{p_i\in B}{\kappa }_i\prod _{p_j\in M}{\lambda }_j\prod _{p_i\in L}{p}_i^2$$

$$h(C)>\prod _{p_i\in B}{\kappa }_i^{\frac{1}{2}}{p}_i^{\frac{e_i}{2}-2}\prod _{p_j\in M}\frac{\sqrt{\sigma (p_j^2)}}{p_j}\prod _{p_i\in L}{p}_i^{e_i-1}.$$

$$p_i={\lambda }_i^{\frac{1}{2}}\prod _{p_i\to p_j,p_j\in N}{p}_j^{\frac{1}{2}}$$

$$q_1=\prod _{q_1^2\to \cdots \to q_k^2\to p^1,p\in L}({\lambda }_{j_1}^{\frac{1}{2}}{\lambda }_{j_2}^{\frac{1}{4}}\cdots {\lambda }_{j_k}^{\frac{1}{2^k}})q_i^{\frac{1}{2^k}}$$

$$\sigma (p_i^{e_i})={\kappa }_i\prod _j{p}_j^{k_{i,j}}={\kappa }_i\prod _{p_i\to p_j,p_j\in N}{p}_j$$

$$\prod _{p_i\in B}\sigma (p_i^{e_i})=\left(\prod _{p_i\in B}{\kappa }_i\right)\prod _{p_j\in M}{\lambda }_j^{s_j}\prod _{p_i\in L}{p}_j^{2s_j},$$

$$s_j=\sum _{\begin{array}{c}{q}_0\to q_1\to \cdots \to q_k=p_j,\\ q_0\in B,q_1,\dots ,q_{k-1}\in N\end{array}}\frac{1}{2^k}=\sum _{\begin{array}{c}{q}_0\to q_1\to \cdots \to q_k=p_j,\\ q_0\in B,q_1,\dots ,q_{k-1}\in N\end{array}}\frac{1}{{\prod }_{l=1}^k{d}_C^{-}(q_l)}.$$

$$\prod _{p_i\in B}\sigma (p_i^{e_i})=\left(\prod _{p_i\in B}{\kappa }_i\right)\left(\prod _{p_j\in M}{\lambda }_j\right)\prod _{p_i\in L}{p}_i^2$$

$$\begin{array}{rl}\prod _{p_i\in C}\frac{\sigma (p_i^{e_i})}{p_i^2}>& \prod _{p_i\in B}{p}_i^{\frac{e_i}{2}-2}\sqrt{\sigma (p_i^{e_i})}\prod _{p_j\in M}\frac{\sigma (p_j^2)}{p_j^2}\prod _{p_i\in L}{p}_i^{e_i-1}\\ \ge & \prod _{p_i\in B}{\kappa }_i^{\frac{1}{2}}{p}_i^{\frac{e_i}{2}-2}\prod _{p_j\in M}{\lambda }_j^{\frac{1}{2}}\frac{\sigma (p_j^2)}{p_j^2}\prod _{p_i\in L}{p}_i^{e_i-1}.\end{array}$$

$$h(C)>\prod _{p_i\in B}{p}_i^{\frac{e_i}{2}-2}.$$

$$h(n)\ge h(\{2,3^2,13,7^e\})h(C(p_2))>h(C(p_2))>1,$$

$$h(\{2,3^2,13\})h(C(p_2))>7^{2}h(\{2,3^2,13\})>1,$$

$$h(n)\ge h(\{2,3^2,13,7^4,2801^2,37,19^4\})>1$$

$$h(n)\ge h(\{2,3^2,13,7^4,2801^2,4933,2467^4\})>1.$$

$$h(n)\ge h(\{2,3^2,13,7^4,2801^2,4933^2,193,97^4\})>1,$$

$$h(n)\ge h(\{2,3^4,11^2,7^4,19\})>1,$$

$$h(n)\ge h(7^6)h(C(L))>h(\{3^2,13^2,61^2,97,7^6\})>1,$$

$$\begin{array}{rl}h(n)\ge & h(\{3^2,13^2,61^2,97\})h(C(p_2)\cup \{7^{10}\})\\ >& 7^{3}h(\{3^2,13^2,61^2,97\})>1.\end{array}$$

$$h(n)\ge h(\{97,7^6\})h(C(p_2))>h(\{97,7^6\})>1.$$

$$h(n)\ge h(\{97,7^4,2801^f\}\cup C(p_2))>2801h(\{97,7^4\})>1.$$

$$h(n)\ge h(\{97,7^4,2801^f\})>2801h(\{97,7^4\})>1.$$

$$h(n)\ge h(\{97,7^4,2801^4\}\cup C(p_2))>h(\{97,7^4,2801^4\})>1.$$

Full text

1. Introduction

Let $\sigma(n)$ and $\gamma(n)$ denote the sum of divisors and the product of distinct prime divisors of $n$, called the radical of $n$, respectively. Moreover, let $\omega(n)$ denote the number of distinct prime divisors of $n$. De Koninck [6] posed the problem to prove or disprove that the only solutions of

[TABLE]

are $n=1,1782$.

According to the editorial comment, it is shown that such an integer $n\neq 1,1782$ must be even, have at least four prime factors, be neither square-free nor squarefull, be greater than $10^{9}$ and have no prime factor raised to a power congruent to $3\ \left(\mathrm{mod}\ 4\right)$. Later, further necessary conditions to satisfy (1) have been shown. Broughan, De Koninck, Kátai and Luca [3] showed that, if an integer $n>1$ satisfies (1), then

[TABLE]

where $p_{i}$ are distinct odd primes and $e_{i}$ are positive integers satisfying (a) $p_{1}\equiv 3\ \left(\mathrm{mod}\ 8\right),e_{1}=1$ and the other $e_{i}$’s are even, or (b) $p_{1}\equiv p_{2}\equiv e_{1}\equiv e_{2}\equiv 1\ \left(\mathrm{mod}\ 4\right),\min\{e_{1},e_{2}\}=1$ and the other $e_{i}$’s are even. Moreover, they showed that $\omega(n)\geq 5$ and $n$ cannot be fourth power free.

Broughan, Delbourgo and Zhou [2] showed that $p_{1}\geq 43$ in the case (a), $p_{1}\geq 173$ in the case (b) with $\alpha_{2}>\alpha_{1}=1$ and $n$ must be divisible by the fourth power of an odd prime. Chen and Tong [5] showed that if $n\neq 1,1782$ satisfies (1) with (a), then $n$ is divisible by $3$ and by the fourth powers of at least two odd primes, $p_{1}\geq 1571$, at most two of $p_{i}$’s are greater than $p_{1}$, $e_{i}=2$ for at least two $i$’s and $e_{i}=2$ for any $i$ such that $10p_{i}^{2}\geq p_{1}$. Moreover, they showed that for any $n$ satisfying (1), at least half of the numbers among $e_{i}+1$’s $(0\leq i\leq s)$ must be either primes or prime squares. Tang and Zhou [10] showed that no integer $n=2^{e_{0}}p_{1}p_{2}p_{3}^{4}p_{4}^{4}$ other than $1,1782$ satisfies (1). Furthermore, there exist only finitely many integers of the form (2) satisfying (1) for any given integer $s$. More generally, Luca [8] showed that, if a positive integer $n$ satisfies $\omega(n)=T$ and $\sigma(n)\mid L(\gamma(n))^{K}$ with $K,L$ positive integers, then

[TABLE]

As usual, $p^{e}\mid\mid n$ denotes that $p^{e}\mid n$ but $p^{e+1}\nmid n$. In this paper, we shall give the following new necessary condition for an integer $n$ to satisfy (1).

Theorem 1.1**.**

If an integer $n\neq 1,1782$ of the form (2) satisfies (1), then there exist odd (not necessarily distinct) primes $p,p^{\prime}$ and (not necessarily odd) distinct primes $q_{i}(i=1,2,\ldots,k)$ with $k\leq 3$ such that $p,p^{\prime}\mid\mid n$, $q_{i}^{2}\mid\mid n(i=1,2,\ldots,k)$ and $q_{1}\mid\sigma(p^{2}),q_{i+1}\mid\sigma(q_{i}^{2})(i=1,2,\ldots,k-1),p^{\prime}\mid\sigma(q_{k}^{2})$.

Our idea is based on the following simple observation, which has been used by previous authors mentioned above. For example, consider the special case $e_{i}=1$ only for $i=1$, $q_{1}\mid\sigma(p^{2})$ for two primes $p$ and for each $p$, $p\mid\sigma(q_{i}^{e_{i}})$ with $e_{i}\geq 4$ for two primes $q_{i}$. Now we have $\sigma(q_{i}^{e_{i}})/q_{i}^{2}>\sqrt{\sigma(q_{i}^{e_{i}})}>p^{1/2}>q_{1}^{1/4}$ for each $i$. Hence, $((q_{1}+1)/q_{1}^{2})\prod_{i}\sigma(q_{i}^{e_{i}})/q_{i}^{2}>q_{1}(q_{1}+1)/q_{1}^{2}>1$ and $\sigma(n)/(\gamma(n))^{2}>1$, which contradicts (1). In order to generalize this observation, we introduce a directed multigraph related to prime power divisors of $n$.

In the next section, we introduce some basic terms on directed multigraphs and prove an identity on directed multigraphs. In Section 3, we introduce a certain directed multigraph related to prime power divisors of $n$ satisfying (1) and give the key lemma for our proof as well as some arithmetic preliminaries.

Under our settings described in Sections 2 and 3, we shall prove the following theorem.

Theorem 1.2**.**

Let $n\neq 1,1782$ be an integer of the form (2) satisfying (1) and $L$ be the set of odd prime divisors $q_{i}$’s with $e_{i}=1$. Let $G(n),N=N(L),M=M(L),B=B(L)$ and $C=C(L)$ be directed multigraphs or sets defined in Section 3. Then,

• i)

If $q_{k+1}\rightarrow q_{k}\rightarrow\cdots\rightarrow q_{1}\rightarrow p$ is a path from a vertex $q_{k+1}$ in $B$ to a vertex $p$ in $L$ via vertices in $M$, then $k\leq 3$ and $q_{i}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ for $1\leq i\leq k-1$.

• ii)

$M$* contains at most two primes $\equiv 1\ \left(\mathrm{mod}\ 3\right)$. Furthermore, $\#M\leq 6$ if $\#L=1$ and $\#M\leq 8$ if $\#L=2$.*

• iii)

There exists a path from $q_{i}$ in $L$ to $q_{j}$ in $L$ consisting of vertices $q_{l}\in N$ other than $q_{i},q_{j}$, where $q_{i}$ and $q_{j}$ may be the same prime.

Now Theorem 1.1 is an arithmetic translation of iii) of Theorem 1.2. In Sections 4 and 5, we prove that the directed multigraph related to prime power divisors of $n$ defined in Section 3 cannot have some forms, which yields iii) of Theorem 1.2. Other statements of Theorem 1.2 easily follow from an elementary divisibility property of values of $\sigma(p^{2})$ with $p$ prime.

2. An identity on directed multigraphs

Before stating our result on directed multigraphs, we would like to introduce some basic terms on directed multigraphs according to [1] with some modifications. A directed multigraph $G=(V,A)$ consists of a set $V$ of elements called vertices and a multiset $A$, where an element may be contained more than once, of ordered pairs of distinct elements in $V$ called arcs. $V=V(G)$ and $A=A(G)$ are called the vertex set and the arc set of $G$ respectively. For an arc $(u,v)$ in $A$, which we call an arc from $u$ to $v$, the former vertex $u$ and the latter vertex $v$ are called its tail and its head respectively. We often write $u\rightarrow v$ if $(u,v)\in A$ and $u\overset{k}{\rightarrow}v$ if $(u,v)\in A$ exactly $k$ times.

The subgraph of $G=(V,A)$ spanned by a given set of vertices $S\subset V$ is the directed multigraph whose vertex set is $S$ and whose arc set consists of all arcs in $A$ whose tail and head both belong to $S$.

A walk $(a_{1},a_{2},\ldots,a_{k})$ of length $k$ is a sequence of arcs $a_{i}=(u_{i},v_{i})(i=1,2,\ldots,k)$ such that $v_{i}=u_{i+1}$ for all $i=1,2,\ldots,k-1$. A walk $(a_{1},a_{2},\ldots,a_{k})$ with $a_{i}=(u_{i},u_{i+1})(i=1,2,\ldots,k)$ is called a path if $u_{1},u_{2},\ldots,u_{k+1}$ are all distinct and a cycle if $u_{1},u_{2},\ldots,u_{k}$ are all distinct and $u_{1}=u_{k+1}$. A walk $(a_{1},a_{2},\ldots,a_{k})$ with $a_{i}=(u_{i},u_{i+1})(i=1,2,\ldots,k)$ is often written as $u_{1}\rightarrow u_{2}\rightarrow\cdots\rightarrow u_{k+1}$. A directed multigraph $G=(V,A)$ is called acyclic if $A$ contains no cycle.

The out-degree $d^{+}(v)=d_{G}^{+}(v)$ and the in-degree $d^{-}(v)=d_{G}^{-}(v)$ of the vertex $v$ are the number of arcs from $v$ and to $v$ respectively counted with multiplicity. A vertex $v$ is called a sink if $d^{+}(v)=0$ and a source if $d^{-}(v)=0$. $S(G)$ denotes the set of sources of the directed multigraph $G$.

Now we would like to state our identity.

Lemma 2.1**.**

Let $G$ be a directed acyclic multigraph. Then, for any vertex $v_{0}$ of $G$ with $d^{-}(v_{0})>0$,

[TABLE]

Proof.

If $G$ consists of only one sink $v_{0}$ and sources $u_{1},u_{2},\ldots,u_{l}$ with arcs $(u_{i},v_{0})$, then (3) is clear.

For any fixed vertices $v_{0},v_{1},\ldots,v_{k-1}$ such that $v_{k-1}\rightarrow v_{k-2}\rightarrow\cdots\rightarrow v_{0}$ and any vertex $w\rightarrow v_{k-1}$ is a source in $G$, we have

[TABLE]

Thus, setting $H$ to be the directed multigraph obtained from $G$ by eliminating all arcs to $v_{k-1}$, we have

[TABLE]

Since $G$ is acyclic, this descent argument eventually reduces $G$ to a directed multigraph $(V,A)$ with $V=\{v_{0},u_{1},u_{2},\ldots,u_{l}\}$ and $A=\{(u_{i},v_{0}),i=1,\ldots,l\}$. Now the lemma follows by induction. ∎

3. A directed multigraph related to divisors of an integer

Let $n$ be a positive integer greater than one. We define the directed multigraph $G=G(n)$ arising from $n$ by setting its vertex set to be the set of primes dividing $n\sigma(n)$ and each arc $p\overset{k}{\rightarrow}q$ to be of multiplicity $k$ if $q^{k}\mid\mid\sigma(p^{e})$ for the exponent $e$ with $p^{e}\mid\mid n$. For convenience, we write $p^{e}\rightarrow q^{f}$ if $p\rightarrow q$ and $p^{e},q^{f}\mid\mid n$ and $p^{e}\in S$ if $p^{e}\mid\mid n$ and $p$ belongs to a set $S$ of vertices.

For a set $S$ of vertices $w_{1},w_{2},\ldots,w_{k}$ of $G$, we define their $2$-incomponent $N(S)$ to be the subgraph of $G$ consisting $w_{1},w_{2},\ldots,w_{k}$ themselves and the vertices $w$ such that there exists a path $v^{2}\rightarrow v_{1}^{2}\rightarrow\cdots\rightarrow v_{l}^{2}\rightarrow w_{i}$ to some vertex $w_{i}$, their $2$-boundary $B(S)$ by the set of vertices $v\not\in N(S)$ from which there exists an edge to some vertex in $N(S)$ and their $2$-closure $C(S)$ by the subgraph whose vertex set is $N(S)\cup B(S)$ and whose arc set consists of all edges in $B(S)$ and all arcs from $N(S)$ to $B(S)$. For convenience, we simply write $N(w)$ for $N(\{w\})$ and so on. Moreover, we put $p_{0}=2$ and $M(S)=N(S)\backslash S$. We note that $C(S)$ may contain $p_{0}=2$.

Now Theorem 1.1 can be restated as in iii) of Theorem 1.2.

For a set $S$ of prime powers, we define $h(S)=\prod_{p^{e}\in S}\sigma(p^{e})/p^{2}$. Clearly, we have $h(S_{0})=\sigma(n)/(\gamma(n))^{2}$ for the set $S_{0}$ of all prime-power divisors of $n$. For convenience, we write $h(p^{e})=h(\{p^{e}\})$ for a prime power $p^{e}$ and $h(n)=h(S_{0})$ for the set $S_{0}$ mentioned above.

We clearly have the following lemma.

Lemma 3.1**.**

We have $h(m)\geq 1$ for any positive integer $m$ with the equality just when $m=1$. If $m_{1}$ divides $m_{2}$, then $h(m_{1})\leq h(m_{2})$. Furthermore, if $S$ and $T$ are disjoint sets of prime-power divisors of $n$, then $h(S\cup T)=h(S)h(T)$.

We also use the following divisibility property of values of the polynomial $x^{2}+x+1$.

Lemma 3.2**.**

If $m$ is an integer and a prime $p$ divides $m^{2}+m+1$, then $p=3$ or $p\equiv 1\ \left(\mathrm{mod}\ 3\right)$. Furthermore, $3$ divides $m^{2}+m+1$ if and only if $m\equiv 1\ \left(\mathrm{mod}\ 3\right)$.

Proof.

The former is a special case of Theorem 94 of [7]. Indeed, if $p\neq 3$ divides $m^{2}+m+1$, then $m\not\equiv 1\ \left(\mathrm{mod}\ p\right)$ and $m^{3}\equiv 1\ \left(\mathrm{mod}\ p\right)$. Hence, $m\ \left(\mathrm{mod}\ p\right)$ has the multiplicative order $3$ and therefore $p-1$ must be divisible by $3$. The latter can be easily confirmed by calculating modulo $3$. ∎

The following lemma is the key point of our proof of Theorem 1.1.

Lemma 3.3**.**

Let $n$ be an integer of the form (2) satisfying (1) and $L$ be a set of prime power divisors of $n$. We define quantities $\kappa_{i}$ for $p_{i}\in C=C(L)$ and $\lambda_{i}$ for $p_{i}\in M=M(L)$ by

[TABLE]

and

[TABLE]

where $\kappa_{i},\lambda_{i}$ are integers not divisible by any prime in $N(L)$.

If $N=N(L)$ is acyclic and any element of $L$ is a sink of $N$, then

[TABLE]

and

[TABLE]

Proof.

We see that

[TABLE]

for $p_{i}\in M$. Since we assume that a vertex in $L$ must be a sink in $C=C(L)$, if $P:q_{1}^{2}\rightarrow\cdots\rightarrow q_{k}^{2}\rightarrow q_{0}$ is a path in $N$ and a prime $q$ in $L$ occurs in $P$, then $q=q_{0}$. Moreover, by the assumption, $N$ is acyclic. Hence, we iterate (10) to obtain

[TABLE]

for any $q_{1}\in M$, where the $j_{m}$’s $(m=1,2,\ldots,k)$ are indices such that $p_{j_{m}}=q_{m}$.

Moreover, we see that

[TABLE]

for $p_{i}\in B$. Combining (11) and (12), we have

[TABLE]

where, observing that $d_{C}^{-}(p_{i})=d_{G}^{-}(p_{i})=2$ for any $p_{i}\in N$ from (1),

[TABLE]

Since $N$ is acyclic by the assumption, Lemma 2.1 gives that $s_{j}=1$ for all $p_{j}\in N$. Thus we obtain

[TABLE]

and therefore

[TABLE]

Now the lemma immediately follows observing that $\lambda_{j}\geq p_{j}^{2}/\sigma(p_{j}^{2})$ for $p_{j}\in M$. ∎

4. Acyclic cases

In this and the next sections, we assume that $n$ is an integer of the form (2) satisfying (1) and we put $L$ to be the set of odd primes $p_{i}$ with $e_{i}=1$. Thus, $L=\{p_{1},p_{2}\}$ in the case (b) with $e_{1}=e_{2}=1$ and $L=\{p_{1}\}$ in the case (a) and the case (b) with $e_{1}=1<e_{2}$. In this section, we shall show that, $N=N(L)$ must have a cycle or we must have $L=\{p_{1},p_{2}\}$ and $p_{1}\in B(p_{2})$ or $p_{2}\in B(p_{1})$.

Lemma 4.1**.**

If $n$ is divisible by $4$ or $2\times 3^{6}$ or $n$ is divisible by $2$ and $3$ does not belong to $C=C(L)$, then $N=N(L)$ must have a cycle or we must have $L=\{p_{1},p_{2}\}$ and $p_{1}\in B(p_{2})$ or $p_{2}\in B(p_{1})$.

Proof.

Assume that $n$ of the form (2) is divisible by $2^{2}$ or $2\times 3^{6}$ and $N$ is acyclic and, in the case $L=\{p_{1},p_{2}\}$, $p_{1}\not\in B(p_{2})$ and $p_{2}\not\in B(p_{1})$.

We can easily see that any prime $p_{i}$ in $L$ must be a sink in $N$. Indeed, if $p_{i}\in L$ and $p_{i}\rightarrow p_{j}$ for some $p_{j}\in N$ not necessarily distinct from $p_{i}$, then, there exists a path from $p_{i}$ to $p_{j}\in L$ via $N$, which contradicts the assumption. Thus, we can apply Lemma 3.3 and, observing that $\kappa_{i}\geq 1$ for all $p_{i}\in B=B(L)$, we obtain

[TABLE]

If $4=2^{2}$ divides $n$, then, observing that $e_{i}/2\geq 2$ for $p_{i}\in B$, Lemma 3.1 and (17) gives that $h(n)\geq h(C)>1$.

If $2$ divides $n$ and $3$ does not belong to $C$, then, by Lemma 3.1, we have $h(n)\geq h(C\cup\{2,3^{2}\})=h(\{2,3^{2}\})h(C)>(3/4)(13/9)>1$.

If $2\times 3^{6}$ divides $n$ and $3$ belongs to $C$, then (17) yields that $h(C)>3$ and $h(n)\geq(3/4)h(C)>9/4>1$.

Thus, in any case, we have $h(n)>1$ or, equivalently, $\sigma(n)>(\gamma(n))^{2}$, which contradicts to the assumption that $n$ satisfies (1). ∎

Now it suffices to settle two cases: $e_{0}=1$ and $3^{2}\in N(L)$ or $e_{0}=1$ and $3^{4}\in B(L)$.

Lemma 4.2**.**

If $e_{0}=1$ and $3^{2}\in N=N(L)$, then $N$ must have a cycle or we must have $L=\{p_{1},p_{2}\}$ and $p_{1}\in B(p_{2})$ or $p_{2}\in B(p_{1})$.

Proof.

Assume that $3^{2}\in N$, $N$ is acyclic and, in the case (b) with $e_{1}=e_{2}=1$, $p_{1}\not\in B(p_{2})$ and $p_{2}\not\in B(p_{1})$. Since $3^{2}$ belongs to $N$, $3^{2}\rightarrow 13$ also belongs to $N$. If $13\in M=M(L)$, then $3^{2}\rightarrow 13^{2}\rightarrow 3$, which contradicts to the assumption that $N$ is acyclic. Thus, $13^{1}\in L$. Now we may assume that $p_{1}=13$. We see that $p_{2}\in L$ and $p_{2}\equiv 1\ \left(\mathrm{mod}\ 4\right)$ since $p_{1}\equiv 1\ \left(\mathrm{mod}\ 4\right)$. Hence, $13\rightarrow 7^{e}$ divides $N$.

We see that $e\geq 2$ must be even since $2^{3}\mid(7+1)$. If $7^{2}\mid\mid N$, then $13\rightarrow 7^{2}\rightarrow 3^{2}\rightarrow 13$, contrary to the assumption that $N$ is acyclic. Thus, $e\geq 4$.

If $7^{e}\not\in B(p_{2})$, then, applying Lemma 3.3, we have

[TABLE]

which is a contradiction. Similarly, if $L=\{p_{1}\}$, then $h(n)\geq h(\{2,3^{2},13,7^{e}\})$ $>1$, which is a contradiction.

Thus, we may assume that $7^{e}\in B(p_{2})$. If $e\geq 8$, then, Lemma 3.3 gives that

[TABLE]

which is a contradiction again.

Assume that $7^{4}\in B(p_{2})$, which immediately yields that $2801\in N(p_{2})$. If $p_{2}=2801$, then $p_{2}\rightarrow 3^{2}\rightarrow 13=p_{1}$, contrary to the assumption that $p_{2}^{e_{2}}\not\in B(p_{1})$. Thus, $2801^{2}\in N(p_{2})$ and $2801^{2}\rightarrow 37,43,4933$.

If $p_{2}=37$, then $p_{3}=19$ divides $n$. If $p_{2}=4933$, then $p_{3}=2467$ divide $n$. In both cases, if $p_{3}^{2}\mid\mid n$, then $p_{2}\rightarrow p_{3}^{2}\rightarrow 3^{2}\rightarrow 13=p_{1}$, which is impossible. If $p_{3}^{4}\mid n$, then

[TABLE]

or

[TABLE]

Hence, $p_{2}=37$ and $p_{2}=4933$ are both impossible.

If $37^{2}\in N(p_{2})$, then $\sigma(37^{2})=3\times 7\times 67$ and therefore $67\in N(p_{2})$. Since $67\equiv 3\ \left(\mathrm{mod}\ 4\right)$, we have $p_{2}\neq 67$ and $37^{2}\rightarrow 67^{2}$. But this implies that $3^{3}\mid\sigma(2\times 37^{2}\times 67^{2})\mid\sigma(n)$, which is a contradiction.

If $4933^{2}\in N(p_{2})$, then $\sigma(4933^{2})=3\times 127\times 193\times 331$ and therefore $p_{2}=193$, since $p_{3}^{2}\in N(p_{2})$ with $p_{3}=127,193$ or $331$ would imply that $3^{3}\mid\sigma(2\times 4933^{2}\times p_{3}^{2})$, a contradiction. Thus $p_{3}=97$ must divide $n$. If $e_{3}=2$, then $3^{3}\mid\sigma(2\times 4933^{2}\times 97^{2})\mid\sigma(n)$, which is impossible. But, if $e_{3}\geq 4$, then

[TABLE]

which is a contradiction again.

If $43^{2}\in N(p_{2})$, then $\sigma(43^{2})=3\times 631$ and therefore $631\in N(p_{2})$. Since $631\equiv 3\ \left(\mathrm{mod}\ 4\right)$, we must have $631^{2}\in N(p_{2})$ and $3^{3}\mid\sigma(2\times 43^{2}\times 631^{2})\mid\sigma(n)$, which is impossible. Thus we see that $2801^{f}\not\in N(p_{2})$ and therefore $7^{4}\not\in N(p_{2})$.

Now we must have $7^{6}\in B(p_{2})$. $\sigma(7^{6})=29\times 4733$ must divide $n$. It is impossible that $p_{2}=29,4733$ since this would imply that $p_{2}\rightarrow 3^{2}\rightarrow 13=p_{1}$. If $29^{2}\in N(p_{2})$, then, observing that $\sigma(29^{2})=13\times 67$ and $\sigma(67^{2})=3\times 7^{2}\times 31$, we must have $29^{2}\rightarrow 67^{2}\rightarrow 31^{2}$. However, this is impossible since $3^{3}\mid\sigma(2\times 67^{2}\times 31^{2})$.

If $4733^{2}\in N(p_{2})$, then, observing that $4733^{2}+4733+1=22406023\equiv 3\ \left(\mathrm{mod}\ 4\right)$ is prime, we must have $22406023^{2}\in N(p_{2})$. If $22406023^{2}\rightarrow p_{2}$, then $p_{2}=1117$ or $p_{2}=606538249$. However, neither of them can occur since $13^{3}\mid\sigma(3^{2}\times 22406023^{2}\times 1117)$ and $5^{3}\mid\sigma(606538249)$. Hence, we must have $22406023^{2}\rightarrow p_{3}^{2}\in N(p_{2})$ for some prime divisor $p_{3}\neq 3$ of $\sigma(22406023^{2})$. But, this is also impossible since $3^{3}\mid\sigma(2\times 22406023^{2}\times p_{3}^{2})$.

Now we conclude that $7$ cannot divide $N$ and therefore $13$ cannot be in $L$. Hence, $3^{2}$ cannot be in $N(L)$. This proves the lemma. ∎

Lemma 4.3**.**

If $e_{0}=1$ and $3^{4}\in B=B(L)$, then $n=1782$, $N=N(L)$ must have a cycle or we must have $L=\{p_{1},p_{2}\}$ and $p_{1}\in B(p_{2})$ or $p_{2}\in B(p_{1})$.

Proof.

Since $3^{4}\in B$, $p_{1}=11$ or $11^{2}\in N$. If $p_{1}=11$, then $n=2\times 3^{4}\times p_{1}=1782$. We note that if $n=n_{0}$ is a solution of (1), then $n=kn_{0}$ with $k>1$ odd and $\gcd(k,n)=1$ can never be a solution of (1). Indeed, $h(n_{0})=h(kn_{0})=1$, then $h(k)=1$. However, this is impossible since $n=1$ is the only odd solution of (1).

Now we may assume that $11^{2}\in N$. If $p_{3}^{2}\in N$ with $p_{3}=7$ or $19$, then $3^{4}\rightarrow 11^{2}\rightarrow p_{3}^{2}\rightarrow 3^{4}$ in $N$, contrary to the assumption. Thus, we must have $p_{1}=19$ and $7^{4}\mid n$. Since $p_{1}\equiv 3\ \left(\mathrm{mod}\ 8\right)$, we must have $L=\{p_{1}\}$. Hence,

[TABLE]

which is impossible again. ∎

5. Cyclic cases

In the previous section, we showed that, if an integer $n$ of the form (2) satisfies (1) and $L$ is the set of odd primes $p_{i}$ with $e_{i}=1$, then $N(L)$ must be cyclic or we must have $L=\{p_{1},p_{2}\}$ and $p_{1}\in B(p_{2})$ or $p_{2}\in B(p_{1})$. In this section, we shall show that $M(L)$ must be acyclic and then complete the proof of Theorems 1.1 and 1.2. We begin by showing that $M=M(L)$ cannot contain a cycle of length $\geq 3$.

Lemma 5.1**.**

Assume that for there exists no arc $p_{i}\rightarrow p_{j}$ from $p_{i}\in L$ to $p_{j}\in N(L)$. Then $M=M(L)$ cannot contain a cycle of length $\geq 3$.

Proof.

Assume that $q_{i}(i=1,2,\ldots,l)$ is a cycle of length $l\geq 3$. We see that $q_{i}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ for all $i$ except possibly one index $j$, for which $q_{j}=3$. We must have $l=3$ and $q_{j}=3$ for some $j$ since otherwise we must have $q_{i}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ for at least three $i$’s by Lemma 3.1 and $3^{3}\mid\prod_{j}\sigma(q_{i}^{2})\mid n$, which is a contradiction.

Now we see that $3^{2}\rightarrow 13^{2}\rightarrow 61^{2}\rightarrow 3^{2}$ is a cycle in $M$ and $p_{1}=97\in L$. Hence, $97\rightarrow 7^{e}$ must divide $n$ and, observing that no more prime $p_{i}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ can satisfy $p_{i}^{2}\mid\mid N$ again, $e\geq 4$ must be even. Moreover, we must have $e_{0}\geq 2$ since $3^{3}\mid\sigma(2\times 13^{2}\times 61^{2})$.

If $L=\{p_{1}\}$ and $7^{6}$ divides $n$, then

[TABLE]

which is a contradiction.

If $L=\{p_{1},p_{2}\}$ and $7^{10}$ divides $n$, then, since $N(p_{2})$ is acyclic, Lemma 3.3 gives

[TABLE]

Now we must have $e=4,6$ or $8$. We can never have $97\rightarrow 7^{8}$ since $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 7^{8})$. In both cases $e=4$ and $e=6$, we have a contradiction that $p^{3}\mid\sigma(n)=(\gamma(n))^{2}$ for some prime $p$ or $h(n)>1$ as follows:

• A.

If $97\rightarrow 7^{6}$, then $\sigma(7^{6})=29\times 4733$ divides $\sigma(n)$. Moreover, we have $L=\{p_{1},p_{2}\}$ with $7^{6}\in B(p_{2})$ or $L=\{p_{1},p_{2}\}$ with $7^{6}\not\in B(p_{2})$ since it is impossible that $L=\{p_{1}\}$ and $7^{6}\mid n$ as seen above.

• A1.

If $7^{6}\in B(p_{2})$, then $p_{2}=29$ or $4733$ or $7^{6}\rightarrow p_{3}^{2}\in N(p_{2})$ with $p_{3}=29$ or $4733$.

• A1a.

If $p_{2}=29$ or $4733$, then $3^{3}\mid\sigma(13^{2}\times 61^{2}\times p_{2})$.

• A1b.

We cannot have $7^{6}\rightarrow 29^{2}$ since $13^{3}\mid\sigma(3^{2}\times 61^{2}\times 29^{2})$.

• A1c.

If $7^{6}\rightarrow 4733^{2}$ and $4733^{2}\in N(p_{2})$, then $p_{2}=22406023$ or $4733^{2}\rightarrow 22406023^{2}$. Since $22406023\equiv 3\ \left(\mathrm{mod}\ 4\right)$, we must have $4733^{2}\rightarrow 22406023^{2}$ and therefore $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 22406023^{2})$.

• A2.

If $L=\{p_{1},p_{2}\}$ and $7^{6}\not\in B(p_{2})$, then, $N(p_{2})$ has no cycle by the assumption and Lemma 3.3 gives

[TABLE]

• B.

If $97\rightarrow 7^{4}$, then $7^{4}\rightarrow 2801^{f}$ for some integer $f>0$.

• B1.

If $f\equiv 1\ \left(\mathrm{mod}\ 4\right)$, then $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 2801)\mid\sigma(n)$.

• B2.

If $f\geq 6$ and $L=\{p_{1},p_{2}\}$, then

[TABLE]

• B3.

If $f\geq 4$ and $L=\{p_{1}\}$, then

[TABLE]

• B4.

If $f=4$ and $2801^{4}\not\in B(p_{2})$, then

[TABLE]

• B5.

If $f=4,L=\{p_{1},p_{2}\}$ and $2801^{4}\in B(p_{2})$, then $q\in N(p_{2})$ with $q=5$, $1956611$ or $6294091$.

• B5a.

If $p_{2}=5$, then $3^{3}\mid\sigma(13^{2}\times 61^{2}\times p_{2})\mid\sigma(n)$.

• B5b.

If $5^{2}\in N(p_{2})$, then $5^{2}\rightarrow 31^{2}$ but $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 31^{2})\mid\sigma(n)$.

• B5c.

We cannot have $6294091^{2}\in N(p_{2})$ since $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 6294091^{2})$.

• B5d.

If $1956611^{2}\in N(p_{2})$, then $\sigma(1956611^{2})=277\times 95479\times 144751$. If $1956611^{2}\rightarrow p_{3}^{2}$ with $p_{3}=277$, $95479$ or $144751$, then $3^{3}\mid\sigma(13^{2}\times 61^{2}\times p_{3}^{2})\mid\sigma(n)$. If $p_{2}=277$, then $h(n)\geq h(\{97,7^{4},2801^{4},277\})>1$.

• B6.

If $f=2$, then $\sigma(2801^{2})=37\times 43\times 4933$ divides $\sigma(n)$.

• B6a.

We cannot have $2801^{2}\rightarrow 43^{2}$ since $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 43^{2})$.

• B6b.

If $2801^{2}\rightarrow 43^{e_{3}},e_{3}\geq 6$, then

[TABLE]

• B6c.

If $2801^{2}\rightarrow 43^{4}$ and $43^{4}\not\in B(p_{2})$, then

[TABLE]

• B6d.

If $2801^{2}\rightarrow 43^{4}$ and $43^{4}\in B(p_{2})$, then $43^{4}\rightarrow 3500201^{1}$ or $43^{4}\rightarrow 3500202^{2}$. If $43^{4}\rightarrow 3500201^{1}$, then $3^{3}\mid\sigma(13^{2}\times 61^{2}\times 3500201)\mid\sigma(n)$. If $43^{4}\rightarrow 3500201^{2}$, then $\sigma(3500201^{2})$ $=13$ $\times 139$ $\times 28411$ $\times 238639$. Since $q\equiv 3\ \left(\mathrm{mod}\ 4\right)$ for $q=139$, $28411$ and $238639$, $q^{2}\in N(q_{2})$ and $3^{3}\mid\sigma(13^{2}\times 61^{2}\times q^{2})\mid\sigma(n)$.

Thus we have a contradiction in any case. This yields that $3^{2}\rightarrow 13^{2}\rightarrow 61^{2}\rightarrow 97$ is impossible. Hence, we conclude that $M=M(L)$ cannot contain a cycle of length $\geq 3$, as stated in the lemma. ∎

Now a cycle in $M(L)$ must be of the form $p_{i}^{2}\leftrightarrow p_{j}^{2}$. We may assume that $p_{r}^{2}\leftrightarrow p_{r+1}^{2}$ for some $r$. In other words, we must have $p_{r}\mid\sigma(p_{r+1}^{2})$ and $p_{r+1}\mid\sigma(p_{r}^{2})$ for some primes $p_{r},p_{r+1}\in M(L)$.

Lemma 2.6 of [5] shows that such $p_{r},p_{r+1}$ must be two consecutive terms of the binary recurrent sequence described in A101368 of OEIS. This had already been proved by Mills [9] and Chao [4]. However, this fact is not needed in our argument. We only use the fact that, if $p_{r+1}>p_{r}>3$ and $p_{r}\leftrightarrow p_{r+1}$, then $p_{r}\equiv p_{r+1}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ by Lemma 3.2.

We begin by proving that, we cannot have $p_{r}\leftrightarrow p_{r+1}$ if $p_{r+1}>p_{r}>3$.

Lemma 5.2**.**

Assume that for there exists no arc $p_{i}\rightarrow p_{j}$ from $p_{i}\in L$ to $p_{j}\in N(L)$. If $M=M(L)$ contains a cycle $p_{r}^{2}\leftrightarrow p_{r+1}^{2}$ of length two with $p_{r+1}>p_{r}$, then $(p_{r},p_{r+1})=(3,13)$.

Proof.

We may assume that $p_{r},p_{r+1}\in N(p_{1})$. Hence, there exists a vertex $q\in N(p_{1})$ such that $p_{r}\rightarrow q$ or $p_{r+1}\rightarrow q$. However, if $q\in M$, then, since $q\equiv p_{r+1}\equiv 1\ \left(\mathrm{mod}\ 3\right)$, we must have $3^{3}\mid\sigma(q^{2}p_{r}^{2}p_{r+1}^{2})\mid\sigma(n)$, which is a contradiction. Thus, we must have $q\in L$.

Now we obtain a directed multigraph $F$ by eliminating the arcs $p_{r}\leftrightarrow p_{r+1}$ and $p_{r}$ or $p_{r+1}\rightarrow p_{i}$ with $p_{i}\in L$ from $C=C(L)$. Then $F$ has two more sinks $p_{r},p_{r+1}$ as well as sinks in $C(L)$.

Proceeding as in the proof of Lemma 3.3, we have

[TABLE]

where

[TABLE]

Let $f_{i}$ be the exponent $p_{i}^{f_{i}}\mid\mid\sigma(p_{r}^{2}p_{r+1}^{2})$ for $p_{i}\in L$. We observe that $d_{F}^{-}(p_{i})=2-f_{i}$ for $p_{i}\in L$, $d_{F}^{-}(p_{r})=d_{F}^{-}(p_{r+1})=1$ and $d_{F}^{-}(p_{j})=2$ for any other vertex $p_{j}$ in $N$. Hence,

[TABLE]

where $t_{j}=(2-f_{i})/2$ for $p_{j}\in L$, $1/2$ for $j=r,r+1$ and $t_{j}=1$ for any other $j$ such that $p_{j}\in N$. By Lemma 2.1, we have $s_{j}=t_{j}$ for any $j$ such that $p_{j}\in N$ and, as in Lemma 3.3,

[TABLE]

If $p_{r}>3$, then we have $p_{r}\equiv p_{r+1}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ and $p_{1}^{f_{1}}p_{2}^{f_{2}}\leq$

$\sigma(p_{r}^{2}p_{r+1}^{2})/(9p_{r}p_{r+1})$. Moreover, we observe that $e_{0}\geq 2$ since $3^{3}\mid\sigma(2p_{r}^{2}p_{r+1}^{2})$. Hence, we must have

[TABLE]

which is a contradiction. Thus, we must have $(p_{r},p_{r+1})=(3,13)$. ∎

Now the only remaining case is $3^{2}\leftrightarrow 13^{2}\rightarrow 61^{1}$.

Lemma 5.3**.**

Assume that there exists no arc $p_{i}\rightarrow p_{j}$ from $p_{i}\in L$ to $p_{j}\in N=N(L)$. Then, $3^{2}\leftrightarrow 13^{2}\rightarrow 61^{1}$ is impossible.

Proof.

Assume that $3^{2}\leftrightarrow 13^{2}\rightarrow 61^{1}$. Then we immediately have $L=\{61\}$ or $L=\{61,p_{2}\}$ with $p_{2}\equiv 1\ \left(\mathrm{mod}\ 4\right)$. It is also clear that $61^{1}\rightarrow 31^{e_{3}}$.

If $e_{3}\geq 8$, then Lemma 3.3 gives

[TABLE]

If $e_{3}\geq 4$ and $L=\{p_{1}\}$, then clearly we have

[TABLE]

If $e_{3}\geq 4$, $L=\{p_{1},p_{2}\}$ and $31^{e_{3}}\not\in B(p_{2})$, then Lemma 3.3 gives

[TABLE]

Thus, in these three cases, we are led to $h(n)>1$, which is a contradiction. Hence, we must have (I) $L=\{p_{1},p_{2}\}$, $e_{3}\in\{4,6\}$ and $31^{e_{3}}\in B(p_{2})$ or (II) $e_{3}=2$. In both cases (I) and (II), we have a contradiction that $p^{3}\mid\sigma(n)=(\gamma(n))^{2}$ for some prime $p$ or $h(n)>1$ as follows:

• I. A.

If $31^{6}\in B(p_{2})$, then $p_{2}=917087137$ or $917087137^{2}\in N(p_{2})$.

• I. A1.

In the case $p_{2}=917087137$, we observe that $p_{4}^{e_{4}}\rightarrow p_{2}$ for a prime $p_{4}\neq 31$.

• I. A1a.

If $e_{4}=2$, then $p_{4}\geq 20612597323$ and, since $3^{2},13^{2},61^{1},31^{6},p_{2}^{1}\not\in C(p_{4})$ (we observe that $p_{2}^{1}\in C(p_{4})$ implies that $N(p_{2})$ must contain a cycle $p_{2}\rightarrow\cdots\rightarrow p_{4}^{2}\rightarrow p_{2}$), Lemma 3.3 yields that

[TABLE]

• I. A1b.

If $e_{4}>2$, then $h(C(p_{2}))>31p_{4}>31^{2}$ by Lemma 3.3 and therefore

[TABLE]

• I. A. 2.

If $917087137^{2}\in N(p_{2})$, then, since any prime factor of $\sigma(p_{4}^{2})$ is $\equiv 3\ \left(\mathrm{mod}\ 4\right)$, we must have $p_{4}^{2}\rightarrow p_{5}^{2}$ with $p_{5}=43,4447,38647$ or $38533987$, which is impossible since $3^{3}\mid\sigma(13^{2}p_{4}^{2}p_{5}^{2})$.

• I. B.

If $31^{4}\in B(p_{2})$, then one of $5,5^{2},11^{2},17351^{2}$ must belong to $N(p_{2})$.

• I. B1.

If $p_{2}=5$, then $h(n)\geq h(\{2,61,31^{4},5\})>1$, a contradiction.

• I. B2.

We cannot have $5^{2}\in N(p_{2})$ since $\sigma(5^{2})=31\in B(p_{2})$.

• I. B3.

If $11^{2}\in N(p_{2})$, then $7^{2}\in N(p_{2})$ or $19^{2}\in N(p_{2})$. Since $\sigma(7^{2})=3\times 19$, we have $19^{2}\in N(p_{2})$ in any case. Now we must have $19^{2}\rightarrow 127^{2}\in N(p_{2})$. Thus, $3^{3}\mid\sigma(13^{2}\times 19^{2}\times 127^{2})\mid\sigma(n)$, a contradiction.

• I. B4.

If $17351^{2}\in N(p_{2})$, then $1063^{2}\in N(p_{2})$ or $21787^{2}\in N(p_{2})$.

• I. B4a.

If $1063^{2}\in N(p_{2})$, then we must have $1063^{2}\rightarrow 377011^{2}\in N(p_{2})$ and $3^{3}\mid\sigma(13^{2}\times 1063^{2}\times 377011^{2})$, which is a contradiction.

• I. B4b.

If $21787^{2}\in N(p_{2})$, then $p_{2}=5104249$ or $5104249^{2}\in N(p_{2})$. Neither of them is possible since $5^{3}\mid(5104249+1)$ and $3^{3}\mid\sigma(13^{2}\times 21787^{2}\times 5104249^{2})$.

• II.

If $61\rightarrow 31^{2}$, then we must have $31^{2}\rightarrow 331^{e_{3}}$ for some $e_{3}$. Since $331\equiv 3\ \left(\mathrm{mod}\ 4\right)$, $p_{2}\neq 331$ and $e_{3}$ must be even.

• II. 1.

$e_{3}=2$ is impossible since $3^{3}\mid\sigma(13^{2}\times 31^{2}\times 331^{2})$.

• II. 2.

If $e_{3}\geq 6$, $L=\{p_{1},p_{2}\}$ and $331^{e_{3}}\in B(p_{2})$, then Lemma 3.3 gives

[TABLE]

• II. 3.

If $e_{3}\geq 4$ and $L=\{p_{1}\}$, then Lemma 3.3 gives

[TABLE]

• II. 4.

If $e_{3}\geq 4$, $L=\{p_{1},p_{2}\}$ and $331^{e_{3}}\not\in B(p_{2})$, then Lemma 3.3 gives

[TABLE]

• II. 5.

If $e_{3}=4$, $L=\{p_{1},p_{2}\}$ and $331^{4}\in B(p_{2})$, then $p_{2}=5,37861,63601$ or $331^{4}\rightarrow p_{4}^{2}\in N(p_{2})$ with $p_{4}=37861$ or $63601$. (we see that since $\sigma(5^{2})=31$, we cannot have $5^{2}\in N(p_{2})$).

• II. 5a.

$331^{4}\rightarrow p_{4}^{2}\in N(p_{2})$ is impossible since $3^{3}\mid\sigma(13^{2}\times 31^{2}\times p_{4}^{2})$.

• II. 5b.

Assume that $p_{2}=5$, $37861$ or $63601$. Since $L=\{p_{1},p_{2}\}$ with $p_{1}=61$, we must have $37861\not\in L$ or $63601\not\in L$. Thus, we see that $331^{4}\rightarrow p_{4}^{e_{4}}$ with $p_{4}=37861$ or $63601$ and $e_{4}\geq 4$. Hence,

[TABLE]

a contradiction again.

Thus we have a contradiction in any case. This shows that $3^{2}\leftrightarrow 13^{2}\rightarrow 61$ is impossible, as desired. ∎

Now we can easily prove Theorem 1.1. Let $n$ be an integer of the form (2) satisfying (1) and $L$ be the set of odd primes $p_{i}$ such that $p_{i}\mid\mid n$. If there exists no path between two vertices in $L$, then, by Lemmas 4.1, 4.2 and 4.3, $N(L)$ must have a cycle but, by Lemmas 5.1, 5.2 and 5.3, $M(L)$ cannot have a cycle. Hence, $G(n)$ must have a path between two vertices in $L$ or a cycle in $N(L)$ containing a vertex in $L$. This proves iii) of Theorem 1.2 and therefore Theorem 1.1.

The remaining statements of Theorem 1.2 can be easily deduced from Lemma 3.2. Let $g_{1}$ and $g_{2}$ be the number of primes $\equiv 1\ \left(\mathrm{mod}\ 3\right)$ and $\not\equiv 1\ \left(\mathrm{mod}\ 3\right)$ in $M$ respectively. i) and the former statement of ii) immediately follow from Lemma 3.2 and the fact that $3^{3}\nmid(\gamma(n))^{2}=\sigma(n)$. Thus, $g_{1}\leq 2$. If $p_{i}$ is a prime $\not\equiv 1\ \left(\mathrm{mod}\ 3\right)$ in $M$, then $p_{i}^{2}\rightarrow p_{j}^{2}$ for some prime $p_{j}\equiv 1\ \left(\mathrm{mod}\ 3\right)$ in $M$ or $p_{i}^{2}\rightarrow p_{l}$ for some prime $p_{l}\in L$. Hence, we obtain $g_{1}+g_{2}\leq 2(g_{1}+\#L)$ and $g_{2}\leq g_{1}+2\#L\leq 2(1+\#L)$. Now the latter statement of ii) follows. This completes the proof of our theorems.