All Questions
Tagged with divisor-sum factoring
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If $s = 4m - 3$, then $\sigma(p^s) = (1 + p^{2m-1})(1 + p + \ldots + p^{2m-2})$. Is there a similar factorization for $\sigma(q^t)$ when $t=4n$?
Let $p,q$ be (odd) primes, and let $m,n,s,t$ be positive integers.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
M. A. Nyblom showed that, if $s = 4m - 3$...
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Does there exist a nontrivial prime power $q^k$ such that $\sigma(n^2)/n = q^k$ for some $n$?
Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$.
My question in the present post is closely related to this one in MO:
QUESTION
Does there exist a nontrivial ...
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A proposed unconditional factor-chain proof for the inequality $I(n) > \frac{3}{2}$, where $q^k n^2$ is an odd perfect number with special prime $q$
Let $N = q^k n^2$ be a hypothetical odd perfect number given in Eulerian form. (That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.)
Denote the ...
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Will it be possible to compute a factored expression for $n^2 - q^k$, if $q^k n^2$ is an odd perfect number with special prime $q$?
In what follows, we denote the classical sum of divisors of the positive integer $x$ by
$$\sigma(x)=\sigma_1(x)=\sum_{d \mid x}{d},$$
and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
If $N$ is ...
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On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part V
(Preamble: This post is an offshoot of this MSE question and this MO question.)
My primary aim in this post is to compute a (hopefully factorable) expression for the quantity $n^2 - q^k$, if $N = q^k ...
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If $p^k m^2$ is an odd perfect number with special prime $p$, then what is wrong about the following factor chain approach to proving $p \neq 5$?
Suppose that $n = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
That $n$ is perfect essentially means that
$$\sigma(p^k)\sigma(m^...
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2
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Tight Bounds on the Sum and Difference of Divisors of RSA Challenge Numbers
The title says it all: given that $n$ of a given length $d$ is a RSA Challenge Number where $n=pq$, where $p$ and $q$ are two primes of length $d/2$. My question is, knowing how the RSA numbers are ...
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How factorize a lot of "little" numbers?
Several fast algorithms are available to factorize big numbers, but what is the best algorithm to factorize a lot of "little" numbers? I need to factorize a lot of odd numbers $< \mathbf{2^{56}}$. ...
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4
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If $q^k n^2$ is an odd perfect number with Euler prime $q$, can $q=17$ hold?
Note: This question is an offshoot of this earlier MSE post.
If $N$ is odd and $\sigma(N)=2N$ where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function, then $N$ is said to be an odd ...
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What is the sum of all positive even divisors of 1000?
I know similar questions and answers have been posted here, but I don't understand the answers. Can anyone show me how to solve this problem in a simple way? This is a math problem for 8th grade ...
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What is the complete (polynomial) factorization of $\sigma(p^k)$, where $p$ is prime with $p \equiv k \equiv 1 \pmod 4$?
The title says it all.
What is the complete (polynomial) factorization of $\sigma(p^k)$, where $p$ is prime with $p \equiv k \equiv 1 \pmod 4$?
Here, $\sigma = \sigma_{1}$ is the classical sum-of-...
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Sum of factors of multiplication of different numbers
Given $N$ numbers $n_i$ such that $\forall i \le N, n_i$ $\le10^9$, is there a method to calculate the sum of divisors of their product?
For example, given $\{11,15,17\}$ their product would is $...
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On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$, Part II
(Note: This has been cross-posted to MO.)
A positive integer $N$ is said to be perfect if $\sigma(N) = 2N$, where $\sigma(x)$ is the sum of the divisors of $x$.
An odd perfect number $N$ is said to ...
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