Additive properties of even perfect numbers

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.

In this paper we are dealing with the problem of the existence of two divisors of $(n^2+1)/2$ whose sum is equal to $\delta n+\varepsilon$, in the case when $\delta$ and $\varepsilon$ are even, or more precisely in the case in which $\delta\equiv\varepsilon+2\equiv0$ or $2 \pmod{4}$. We will completely solve the cases $\delta=2, \delta=4$ and $\varepsilon=0$.

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 ...

We shall give some results for an integer divisible by its unitary totient.

In this note, we continue an approach pursued in an earlier paper of the second author and thereby attempt to produce an improved lower bound for the sum $I(q^k) + I(n^2)$, where $q^k n^2$ is an odd perfect number with special prime $q$ and $I(x)$ is the abundancy index of the positive integer $x$. In particular, this yields an upper bound for $k$.

We proved that any even number not less than 6 can be expressed as the sum of two old primes, $2n=p_i+p_j$

Let $A$ be a set of $n$ positive integers. We say that a subset $B$ of $A$ is a divisor of $A$, if the sum of the elements in $B$ divides the sum of the elements in $A$. We are interested in the following extremal problem. For each $n$, what is the maximum number of divisors a set of $n$ positive integers can have? We determine this function exactly for all values of $n$. Moreover, for each $n$ we characterize all sets that achieve the maximum. We also prove results for the $...

Inspired by the question of Graeme L. Cohen and Herman J. J. te Riele , The Authors of \cite{Graeme L} who they investigated a question Given $n$ is there an integer $k$ for which $\sigma^k(n) = 0 \bmod n$? They did this in a $1995$ paper and asserted through computation that the answer was yes for $n \leq 400$,The aim of this paper is to give a negative answer to the reverse question of Graeme L. Cohen and Herman J. J. te Riele such that we shall prove that there is no fixed...

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 $7|N$ ...

We will show the two following results: If there existe an odd perfect number $n$ of prime decomposition $n=p_1^{\alpha_1} \ldots p_k^{\alpha_k}q^\beta$, where the $\alpha_i$ are even, the $\beta$ are odd and $q \equiv 5 \mod 8$. Then there is at least one $p_i$, $1 \leq i \leq k$ that is not a square in $\mathbb{Z}/q\mathbb{Z}$. More precisely there is an odd number of $p_i$ that are not squares in $\mathbb{Z}/q\mathbb{Z}$. If there exist an odd perfect number $n$ of prime d...