All Questions
Tagged with divisor-sum modular-arithmetic
24
questions
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Correctness of $\sigma_s(n) (1-\frac{1}{n^s}) = 0 \pmod{k}$ (divisor sum congruence)
I've played around with some identities and came up with:
$\sigma_s(n)(1-\frac{1}{n^s}) = 0 \pmod{k}$
(for $n$ and $k$ positive integers, and $s$ an integer)
With the conditions that 1) $n$ is ...
4
votes
1
answer
331
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Ratios on bounds of sum of divisors of $8n+5$ and $2n+1$
I am trying to understand if it can be shown that $\frac{S(8n+5)}{S(2n+1)}<1$ is not possible, where $S(n)$ is the sum of divisors of $n$. There seems to be no example that's available till $100$ ...
1
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1
answer
71
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If $p^k m$ is deficient (with $p$ a prime number satisfying $\gcd(p,m)=1$), then what is the sign/residue of $D(p^k)D(m) - 2s(p^k)s(m)$ (modulo $3$)?
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)-x$. ...
1
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1
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71
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Is there an odd $x$ such that $2x^2 \equiv 0 \pmod {\sigma(x)}$ and $\sigma(x^2) \equiv 0 \pmod {\sigma(x) - 1}$?
CONTEXT
This question is a result of considerations stemming from this closely related MO question.
INITIAL QUESTION
My question is as is in the title:
Is there an odd $x$ such that $$2x^2 \equiv 0 \...
1
vote
1
answer
57
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If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).
I tried Googling for the keywords "the theory of odd perfect numbers" and one of the search results that came up was this document, titled ON THE DIVISORS OF THE SUM OF A GEOMETRICAL SERIES ...
3
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1
answer
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Prime dividing the divisors sum of a prime-power
Let $p$ and $q$ be distinct primes and $n \in \mathbb{N}$ such that:
$q\vert \sigma(p^n)=\frac{p^{n+1}-1}{p-1}$ and $q \vert p-1$
How would one go about proving the equivalence to $q \vert n+1$?
It ...
1
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0
answers
37
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What is $\sigma(q^k)$ in terms of $k$ if $q \equiv 5 \pmod 8$?
Denote the sum of divisors of $x \in \mathbb{N}$ by $\sigma(x)$.
Here is my question:
What is $\sigma(q^k)$ in terms of $k$ if $q \equiv 5 \pmod 8$?
I know that if $q \equiv 1 \pmod 8$, then
$$\...
1
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0
answers
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How to simplify $2 q^{\frac{k-1}{2}} n^2 - \sigma(q^{\frac{k-1}{2}})\sigma(n^2)$
Let $k$ be a positive integer satisfying $k \equiv 1 \pmod 4$. Let $x \in \mathbb{N}$. Let $q$ be a prime number.
If
$$\sigma(x) = \sum_{d \mid x}{d}$$
is the classical sum-of-divisors function, ...
1
vote
0
answers
60
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Around the equation $4\,\varphi \left(2^{n-1} \sigma(n)\right) ^2=2^{\sigma(n)}\cdot\varphi(n)^2$
Here for integers $m\geq 1$ I denote the Euler's totient function as $\varphi(m)$ and the sum of positive divisors $\sum_{d\mid m}d$, as $\sigma(m)$.
I know that it is possible to prove the following ...
15
votes
2
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$\sigma(n) \equiv 1 \space \pmod{n}$ if and only if $n$ is prime
For $n>1$, let $\sigma(n)$ denote the sum of all positive integers that evenly divide $n$. Then $\sigma(n) ≡ 1 \space(mod\space n)$ if and only if $n$ is prime.
I've been trying to prove this for ...
1
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6
answers
2k
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Sum of positive divisors if and only if perfect square
let n be a positive odd integer, prove that the sum of the positive divisors of n is odd if and only if n is a perfect square.
I know that based on the prime factorization theory that every integer n ...
0
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1
answer
54
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Mods and Multiplicative functions
Where m is the largest odd factor of n, I am trying to prove that $$\sigma(n)\equiv d(m)\mod{2}$$
Any help is appreciated!
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1
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If $\sigma(n) \equiv 2 \pmod 4$ and $n$ is odd, does this imply that $n = p^k m^2$?
Let $\sigma(n)$ denote the sum of divisors of $n$.
Here is my question:
If $\sigma(n) \equiv 2 \pmod 4$ and $n$ is odd, does this imply that $n = p^k m^2$ where $p \equiv k \equiv 1 \pmod 4$ and $\...
0
votes
1
answer
92
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When does $\sigma_{1}(p^{2k}) = q$ have a solution, where $p$ and $q$ are primes, and $k \geq 1$?
Let $\sigma_{1}$ be the classical sum-of-divisors function. For example, $\sigma(12)=1+2+3+4+6+12=28$.
Here is my question:
When does $\sigma_{1}(p^{2k}) = q$ have a solution, where $p$ and $q$ ...
2
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0
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On identities and congruences involving the harmonic mean of odd perfect numbers
I would like to know if my calculations were rights and if you want to deduce other different interesting congruences following my approach.
On assumption (notice that it isn't the usually accepted ...