Solving the Odd Perfect Number Problem: Some Old and New Approaches

Let $N$ be an odd perfect number. Then, Euler proved that there exist some integers $n, \alpha$ and a prime $q$ such that $N = n^{2}q^{\alpha}$, $q \nmid n$, and $q \equiv \alpha \equiv 1 \bmod 4$. In this note, we prove that the ratio $\frac{\sigma(n^{2})}{q^{\alpha}}$ is neither a square nor a square times a single prime unless $\alpha = 1$. It is a direct consequence of a certain property of the Diophantine equation $2ln^{2} = 1+q+ \cdots +q^{\alpha}$, where $l$ denotes on...

Euler showed that if an odd perfect number exists, it must be of the form $N = p^\alpha q_{1}^{2\beta_{1}}$ $\ldots$ $q_{k}^{2\beta_{k}}$, where $p, q_{1}, \ldots, q_k$ are distinct odd primes, $\alpha$, $\beta_{i} \geq 1$, for $1 \leq i \leq k$, with $p \equiv \alpha \equiv 1 \pmod{4}$. In 2005, Evans and Pearlman showed that $N$ is not perfect, if $3|N$ or $7|N$ and each $\beta_{i} \equiv 2 \pmod{5}$. We improve on this result by removing the hypothesis that $3|N$ or $7|N$ ...

Let $n$ and $k$ be positive integers and $\sigma(n)$ the sum of all positive divisors of $n$. We call $n$ an exactly $k$-deficient-perfect number with deficient divisors $d_1, d_2, \ldots, d_k$ if $d_1, d_2, \ldots, d_k$ are distinct proper divisors of $n$ and $\sigma (n)=2n-(d_1+d_2+\ldots + d_k)$. In this article, we show that the only odd exactly $3$-deficient-perfect number with at most two distinct prime factors is $1521=3^2 \cdot 13^2$.

The existence of a perfect odd number is an old open problem of number theory. An Euler's theorem states that if an odd integer $ n $ is perfect, then $ n $ is written as $ n = p ^ rm ^ 2 $, where $ r, m $ are odd numbers, $ p $ is a prime number of the form $ 4 k + 1 $ and $ (p, m) = 1 $, where $ (x, y) $ denotes the greatest common divisor of $ x $ and $ y $. In this article we show that the exponent $ r $, of $ p $, in this equation, is necessarily equal to 1. That is, if ...

Some new results concerning the equation $\sigma(N)=aM, \sigma(M)=bN$ are proved. As a corollary, there are only finitely many odd superperfect numbers with a fixed number of distinct prime factors.

In this paper, we prove the conjecture that if there is an odd perfect number, then there are infinitely many of them.

Let $p^k m^2$ be an odd perfect number with special prime $p$. In this article, we provide an alternative proof for the biconditional that $\sigma(m^2) \equiv 1 \pmod 4$ holds if and only if $p \equiv k \pmod 8$. We then give an application of this result to the case when $\sigma(m^2)/p^k$ is a square.

Let $\sigma(n)$ to be the sum of the positive divisors of $n$. A number is non-deficient if $\sigma(n) \geq 2n$. We establish new lower bounds for the number of distinct prime factors of an odd non-deficient number in terms of its second smallest, third smallest and fourth smallest prime factors. We also obtain tighter bounds for odd perfect numbers. We also discuss the behavior of $\sigma(n!+1)$, $\sigma(2^n+1)$, and related sequences.

Let $q^k n^2$ be an odd perfect number with special prime $q$. Define the GCDs $$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$$ and $$I = \gcd\bigg(n,\sigma(n^2)\bigg).$$ We prove that $G \times H = I^2$. (Note that it is trivial to show that $G \mid I$ and $I \mid H$ both hold.) We then compute expressions for $G, H,$ and $I$ in terms of $\sigma(q^k)/2, n,$ and $\gcd\bigg(\sigma(q^k)/2,n\bigg)$. Afterwards, we prove that if $G = H = I$,...

Let $N$ be an odd perfect number. Let $\omega(N)$ be the number of distinct prime factors of $N$ and let $\Omega(N)$ be the total number of prime factors of $N$. We prove that if $(3,N)=1$, then $ \frac{302}{113}\omega - \frac{286}{113} \leq \Omega. $ If $3|N$, then $\frac{66}{25}\omega-5\leq\Omega.$ This is an improvement on similar prior results by the author which was an improvement of a result of Ochem and Rao. We also establish new lower bounds on $\omega(N)$ in terms ...