It is shown that every 3-perfect number in its prime factorization has the exponent of the number 2 which is greater than 1.
Let us recall some concepts, formulas and notations. The set of all natural numbers \(\{1,2,3,\ldots\}\) will be denoted by \(\mathbb{N}.\) \(\mathbb{R}^+\) is the set of all positive real numbers. We will use the notion \((n,m)\) of greatest common divisor of two numbers \(n,m\in \mathbb{N}.\) If \(n\in \mathbb{N}\) and \(n>1,\) then its factorization is the representation of number \(n\) in the form \(n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s},\) where \(p_i\) is prime, \(p_1<p_2<\ldots<p_s\) and \(\alpha_i\in \mathbb{N}\) for any \(i=\overline{1,s}.\)
A function \(f\) is called arithmetic if it is defined on the set of all natural numbers. An arithmetic function \(f\) is called multiplicative if \(f(1)=1\) and for any numbers \(a,b\in \mathbb{N}\) with \((a, b)=1\) it follows that \(f(ab)=f(a)f(b).\)
The arithmetic function \(\sigma\) is defined for any \(n\in \mathbb{N}\) as the sum of all positive divisors of \(n,\) i.e. \(\sigma(n)=\underset{d|n}{\sum}d.\) It is known that the function \(\sigma\) is multiplicative and if \(n=p_1^{\alpha_1}p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}\) is factorization of the number \(n,\) then \[\sigma(n)=\frac{p_1^{\alpha_1+1}-1}{p_1-1}\cdot\frac{p_2^{\alpha_2+1}-1}{p_2-1} \cdot\ldots\cdot\frac{p_s^{\alpha_s+1}-1}{p_s-1}.\]
In [1] for a natural number \(n\) the multiplicative function \(h(n)=\displaystyle\frac{\sigma(n)}{n}\) and some of its properties are considered. Let us formulate and give a proof of one of them, which we will apply hereafter.
Lemma 1. [1] Let \(p>q\) be odd prime numbers, \(\alpha\in \mathbb{N},\) then \[\label{1} h(p^\alpha)<h(q). \tag{1}\]
Proof.Let \(x,t\in \mathbb{R}^+,\) \(t>1\) we consider the function \(g:\mathbb{R}^+\rightarrow \mathbb{R}^+\) such that \[g(t^x)=\frac{t^{x+1}-1}{t^{x+1}-t^x}.\]
Since the function \(g(t^x)\) is continuous at every point of the set \(\mathbb{R}^+,\) then \(g(t^x)\) is continuous on the entire set \(\mathbb{R}^+.\) Next, we will show that the function \(g(t^x)\) is monotonically increasing on \(\mathbb{R}^+.\) Suppose \(x_1, x_2\in \mathbb{R}^+\) and \(x_1 > x_2.\) Since \(t>1,\) it follows that \(t^{x_1}>t^{x_2}.\) We will show that \(g(t^{x_1})>g(t^{x_2}).\) Let’s assume the contrary, that is, \(g(t^{x_2})\geq g(t^{x_1}).\) Then we have
\[\begin{aligned} \frac{t^{x_2+1}-1}{t^{x_2+1}-t^{x_2}}&\geq \frac{t^{x_1+1}-1}{t^{x_1+1}-t^{x_1}},\\ \left(t^{x_2+1}-1\right)\left(t^{x_1+1}-t^{x_1}\right)&\geq \left(t^{x_1+1}-1\right)\left(t^{x_2+1}-t^{x_2}\right),\\ t^{x_2+1}t^{x_1+1}-t^{x_2+1}t^{x_1}-t^{x_1+1}+t^{x_1} &\geq t^{x_1+1}t^{x_2+1}-t^{x_1+1}t^{x_2}-t^{x_2+1}+t^{x_2},\\ \left(t^{x_1+1}t^{x_2}-t^{x_2+1}t^{x_1}\right)+t^{x_1}\left(1-t\right) &\geq t^{x_2}\left(1-t\right),\\ t^{x_1}t^{x_2}\left(t-t\right)+t^{x_1}\left(1-t\right) &\geq t^{x_2}\left(1-t\right),\ \text{since}\ t>1,\ \text{then}\\ t^{x_1} &\leq t^{x_2}. \end{aligned}\]
The obtained contradiction proves that the function \(g(t^x)\) is monotonically increasing on \(\mathbb{R}^+.\)
Then \[g(t^x)=\frac{t^{x+1}-1}{t^{x+1}-t^x}=\frac{t-\frac{1}{t^x}}{t-1}<\underset{x\rightarrow \infty}{\lim}g(t^x)=\frac{t}{t-1},\] and \(g(p^\alpha)<\frac{p}{p-1}.\)
We obtain the inequality, \[h(p^\alpha)=\frac{p^{\alpha+1}-1}{p^{\alpha+1}-p^\alpha}=g(p^\alpha)<\frac{p}{p-1}=1+\frac{1}{p-1}.\]
Since \(p\) and \(q\) are odd prime numbers and \(p>q,\) then \(p-1\) is an even number such that \(p-1>q.\) Therefore \[1+\frac{1}{p-1}<1+\frac{1}{q}=\frac{q+1}{q}=h(q).\]
Thus \(h(p^\alpha)<h(q).\) ◻
Recall that a natural number n is called perfect if it is equal to the sum of its own divisors, i.e. \[n=\underset{d|n,\,d\geq 1,\,d\neq n}{\sum}d.\]
Let \(n,k\in \mathbb{N}\) such that \(k, n>1.\) A number \(n\) is called \(k\)-perfect if \(\sigma(n)=kn.\) The concept of a perfect number of multiplicity k is equivalent to the concept of a \(k\)-perfect number, the history of research of which can be found, e.g., in [2]. It follows from the definition of \(k\)-perfect number that the concepts of perfect and \(2\)-perfect numbers coincide. For example, 6 and 28 are \(2\)-perfect numbers, and 120 and 672 are \(3\)-perfect numbers. Even perfect numbers were described by Euclid and Euler. Some necessary and sufficient conditions for the existence of odd perfect numbers were found in [3].
Lemma 2. Let \(k,n\in \mathbb{N}\) be such that \(k, n>1\) and \((k, n)=1.\) A number \(n\) is \(k\)-perfect if and only if \(kn\) is \(\sigma(k)\)-perfect.
Proof. Let the number \(n\) be \(k\)-perfect. Let us show that \(kn\) is \(\sigma(k)\)-perfect. Indeed, \(\sigma(kn)=\sigma(k)\sigma(n)=\sigma(k)\cdot kn,\) i.e. \(\sigma(kn)=\sigma(k)\cdot kn.\)
Conversely. Let the number \(kn\) be \(\sigma(k)\)-perfect. Let us show that the number \(n\) is \(k\)-perfect. Indeed, \(\sigma(k)\sigma(n)=\sigma(kn)=\) (because \(kn\) is \(\sigma(k)\)-perfect) \(=\sigma(k)\cdot kn.\) Hence, \(\sigma(k)\sigma(n)=\sigma(k)\cdot kn.\) Since \(\sigma(k)\neq 0\) then, reducing both sides of the last equality by \(\sigma(k),\) we get \(\sigma(n)=kn.\) ◻
Corollary 1. Let odd perfect numbers exist. Then the odd number \(n>1\) is perfect if and only if the number \(2n\) is \(3\)-perfect.
Proof. To prove this corollary, it is sufficient to put \(k=2\) in the formulation of Lemma 2. ◻
Further we will use the following statement proved in [4]. For completeness we will give its proof.
Lemma 3. [4] Let \(n\in \mathbb{N}\) be a \(3\)-perfect number and \(n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}\) be its factorization, then \(s\geq 3.\)
Proof. Suppose the contrary, i.e. let \(s=2\) (the case of \(s=1\) is proved similarly). Then \(n=p_1^{\alpha_1}p_2^{\alpha_2}.\) Consider the expression \[3p_1^{\alpha_1}p_2^{\alpha_2}=\sigma(n)=\frac{p_1^{\alpha_1+1}-1}{p_1-1}\cdot\displaystyle\frac{p_2^{\alpha_2+1}-1}{p_2-1}<\displaystyle\frac{p_1^{\alpha_1+1}}{p_1-1} \cdot\displaystyle\frac{p_2^{\alpha_2+1}}{p_2-1},\] or \[3<\displaystyle\frac{p_1}{p_1-1}\cdot\displaystyle\frac{p_2}{p_2-1}\leq\displaystyle\frac{2}{2-1}\cdot\displaystyle\frac{3}{3-1}=3.\]
The resulting contradiction proves this statement. ◻
Theorem 1. Every even \(3\)-perfect number has in its factorization the exponent of number 2 greater than 1.
Proof. Let \(n=2^{\alpha_1}\cdot p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}\) be its factorization of any even \(3\)-perfect number. Then we have \[3\cdot 2^{\alpha_1}\cdot p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}=\sigma(n)=\left(2^{\alpha_1+1}-1\right)\sigma\left(p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}\right).\]
Hence, \[2^{\alpha_1}=\displaystyle\frac{1}{3}\left(2^{\alpha_1+1}-1\right)\displaystyle\frac{\sigma\left(p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}\right)}{p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}}=\displaystyle\frac{1}{3}\left(2^{\alpha_1+1}-1\right)t,\]where \[\begin{aligned} t&=\displaystyle\frac{\sigma\left(p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}\right)}{p_2^{\alpha_2}\cdot\ldots\cdot p_s^{\alpha_s}}=\overset{s}{\underset{i=2}{\prod}}h(p_i^{\alpha_i}),\ s\geq 3. \end{aligned}\]
Then \[2^{\alpha_1}=\displaystyle\frac{1}{2-\displaystyle\frac{3}{t}}\ .\]
Consider the following cases.
1. Let \(p_2\neq 3.\) Then according to inequality (1) it follows that \(h(p_i^{\alpha_i})<h(3)\) for any \(i=\overline{2,s}.\) Therefore, \(t<h(3)^{s-1},\) where \(s\geq 3.\) Thus, \[2^{\alpha_1}=\displaystyle\frac{1}{2-\displaystyle\frac{3}{t}}>\displaystyle\frac{1}{2-\displaystyle\frac{3}{h(3)^{s-1}}},\ \text{i.\,e.}\] \[\label{4} 2^{\alpha_1}>\displaystyle\frac{1}{2-\displaystyle\frac{3}{h(3)^{s-1}}}\ . \tag{2}\]
By the condition \(\alpha_1\geq 1,\) then \(\alpha_1>0.9.\) Therefore, \(2^{\alpha_1}>2^{0.9}\) Next, we will find at what values of the variable s the right side of inequality (2) will be greater than \(2^{0.9}.\) We have \[\displaystyle\frac{1}{2-\displaystyle\frac{3}{h(3)^{s-1}}}> 2^{0.9}.\]
From this we obtain that \[\label{5} h(3)^{s-1}< \frac{3\cdot 2^{0.9}}{2^{1.9}-1}\approx 2.049. \tag{3}\]
Since \(h(3)=4/3,\) then from inequality (3) we obtain that \(s< 1+\displaystyle\frac{\ln 2.049}{\ln\displaystyle\frac{4}{3}}\approx 3.5.\) Hence, \(s=3.\) Then we have \[2^{\alpha_1}>\displaystyle\frac{1}{2-\displaystyle\frac{3}{\left(\displaystyle\frac{4}{3}\right)^2}}=\displaystyle\frac{16}{5}.\]
So \(\alpha_1>4-\log_25>1.\)
2. Let \(p_2=3.\) Then \[t=\overset{s}{\underset{i=2}{\prod}}h(p_i^{\alpha_i})=h(3^{\alpha_2})\overset{s}{\underset{i=3}{\prod}}h(p_i^{\alpha_i})<h(3^{\alpha_2})h(3)^{s-2},\] (since according to inequality (1) we have \(h(p_i^{\alpha_i})<h(3)\) for any \(i=\overline{3,s}\) ). Thus, \[\label{6} 2^{\alpha_1}=\displaystyle\frac{1}{2-\displaystyle\frac{3}{t}}>\displaystyle\frac{1}{2-\displaystyle\frac{3}{h(3^{\alpha_2})h(3)^{s-2}}},\ s\geq 3. \tag{4}\]
As in case 1, we restrict the right-hand side of inequality (2) from below by a number \(2^{0.9}\), i.e., we have \[\label{7} \displaystyle\frac{1}{2-\displaystyle\frac{3}{h(3^{\alpha_2})h(3)^{s-2}}}> 2^{0.9}. \tag{5}\]
Then from inequality (3) we obtain that \[h(3^{\alpha_2})h(3)^{s-2}< 2.049,\ s\geq 3.\]
Then we have \[\left(1+\displaystyle\frac{1}{3}+\ldots+\displaystyle\frac{1}{3^{\alpha_2}}\right)\left(\displaystyle\frac{4}{3}\right)^{s-2}< 2.049\ \ \text{or}\] \[\left(1+\displaystyle\frac{1}{3}\right)\left(\displaystyle\frac{4}{3}\right)^{s-2}+\left(\displaystyle\frac{1}{3^2}+\ldots+\displaystyle\frac{1}{3^{\alpha_2}}\right) \left(\displaystyle\frac{4}{3}\right)^{s-2}< 2.049.\]
Hence, \[\left(1+\frac{1}{3}\right)\left(\frac{4}{3}\right)^{s-2}< 2.049,\ \text{or}\ \left(\frac{4}{3}\right)^{s-1}< 2.049.\]
Then \(s=3\) (see case 1). Then we have \[\begin{gathered} \begin{split} 2^{\alpha_1}&=\displaystyle\frac{1}{2-\displaystyle\frac{3}{t}}>\displaystyle\frac{1}{2-\displaystyle\frac{3}{h(3^{\alpha_2})h(3)}}= \displaystyle\frac{1}{2-\displaystyle\frac{3}{\displaystyle\frac{(3^{\alpha_2+1}-1)}{3^{\alpha_2}\cdot(3-1)}\cdot \displaystyle\frac{4}{3}}}= \displaystyle\frac{2\cdot 3^{\alpha_2+1}-2}{3^{\alpha_2+1}-4}=\\ &=\displaystyle\frac{2(3^{\alpha_2+1}-4)+6}{3^{\alpha_2+1}-4}= 2+\displaystyle\frac{6}{3^{\alpha_2+1}-4}>2, \end{split} \end{gathered}\] (since \(\alpha_2\geq 1\)).) Therefore, \(2^{\alpha_1}>2,\) i.e. \(\alpha_1>1.\) ◻
Corollary 2. There are no odd perfect numbers.
Proof. Suppose the contrary, i.e. let an odd perfect number \(n\) exists. Then the number \(2n\) is \(3\)-perfect by Corollary 2, which contradicts Theorem 1. ◻
This work does not have any conflicts of interest
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