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JOURNALS // Vestnik Tomskogo Gosudarstvennogo Universiteta. Matematika i Mekhanika // Archive

Vestn. Tomsk. Gos. Univ. Mat. Mekh., 2014 Number 6(32), Pages 5–18 (Mi vtgu423)

MATHEMATICS

On odd perfect numbers

R. Z. Ahmadullin

Bashkir State Pedagogical University named after M. Akmulla, Ufa, Russian Federation

Abstract: A perfect number is a natural number equal to the sum of all its proper divisors (all positive divisors other than the number itself). Perfect numbers form a sequence: $6, 28, 496, 8128, 33550336, 8589869056, 137438691328~\dots$
Let $S=p_1^{a_1}\cdot p_2^{a_2}\cdots p_{n-1}^{a_{n-1}}\cdot p_n^{a_n}$ be a perfect number, where $p_i$ are primes, $a_i$ are some natural numbers, $a_i\geqslant 1$, $i=1, \dots, n$, and $n$ is the number of factors of the number $S$. Then
\begin{equation} \frac{p_1^{1+a_1}-1}{(p_1-1)p_1^{a_1}}\cdot \frac{p_2^{1+a_2}-1}{(p_2-1)^*p_2^{a_2}}\cdots \frac{p_n^{1+a_n}-1}{(p_n-1)^*p_n^{a_n}}=2. \end{equation}

Equation (1) is a Diophantine equation with an indefinite number of unknowns; it contains $2n$ unknowns, the value of $n$ (the number of factors of the number) is not fixed. This equation is equivalent to the two systems:
\begin{equation} \left\{\begin{aligned} p_1=&\frac1{\mathcal{Q}_1-1}\geqslant2, &a_1=\frac{-\ln(\mathcal{Q}_1-p_1(\mathcal{Q}_1-1))}{\ln(p_1)}\geqslant1;\\ & &\dots;\\ p_n=&\frac1{\mathcal{Q}_n-1}\geqslant p(n), & a_n=\frac{-\ln(\mathcal{Q}_n-p_n(\mathcal{Q}_n-1))}{\ln(p_n)}\geqslant1,\\ \end{aligned}\right. \end{equation}
where
$$ \begin{gathered} \mathcal{Q}_i=2\frac{(p_1-1)p_1^{a_1}}{p_1^{1+a_1}-1}\cdots \frac{(p_{i-1}-1)p_{i-1}^{a_{i-1}}}{p_{i-1}^{1+a_{i-1}}-1}\cdot \frac{(p_{i+1}-1)p_{i+1}^{a_{i+1}}}{p_{i+1}^{1+a_{i+1}}-1}\cdots \frac{(p_n-1)p_n^{a_n}}{p_n^{1+a_n}-1}=\\ =2\prod_{j=1}^{n\backslash i}\frac{(p_j-1)p_j^{a_j}}{p_j^{1+a_j}-1};\quad i=1,\dots,n, \end{gathered} $$
and
\begin{equation} \left\{\begin{aligned} \frac{p_1^{1+a_1}-1}{(p_1-1)}=&2^{\delta(a_1,p_1)}\prod_{j=1}^{n\backslash1}p_i^{a^{(1,j)}(a_1,p_1)};\dots;\\ \frac{p_n^{1+a_n}-1}{(p_n-1)}=&2^{\delta(a_n,p_n)}\prod_{j=1}^{n-1}p_i^{a^{(n,j)}(a_n,p_n)};\quad \sum_{j=1}^n a^{(j)}=a_i; \quad i=1,\dots,n, \end{aligned} \right. \end{equation}
where $\delta(a_1,p_1)$ is formally defined as follows:
$$ \delta(a_1,p_1)= \begin{cases} 0, & \text{ if } p_1=2,\\ 0, &\text{ if } p_1\ne2 \text{ and } a_1\text{- even},\\ 1, &\text{ if } p_1\ne2 \text{ and } a_1\text{- odd}. \end{cases} $$

With allowance for the fact that the factorization of natural numbers is determined uniquely, the system of equations (5) is a system of $2n$ equations and $2n$ unknowns (not with $(n^2+n)$ unknowns). The numbers $a^{(i,j)}$ are uniquely determined by a factorization function $F(p_1,a_1,i,j)$ and are considered as parameters.
From the system of equations (2) we obtain the equation
\begin{equation} a=-\frac{\ln\left(q-\frac1{q-1}(q-1)\right)^{-1}}{\ln\frac1{q-1}} \end{equation}
at $2>q>1$. This function has an infinite number of (infinite) left discontinuities of the second kind at the points $q=(l+1)/1$ ($l\in\mathrm{N}$). Hypothetically, beginning from some values of $n$, most of exponents of $a_n$ in system (2) can be equal only to $1$.
It is proved that for a given (fixed) value $n\geqslant3$ there exists only a finite number of odd perfect numbers.

Keywords: odd perfect number, amicable numbers, number theory.

UDC: 511.2

Received: 02.02.2014



© Steklov Math. Inst. of RAS, 2026