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1 vote
0 answers
31 views

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 ...
Niklas's user avatar
  • 341
4 votes
1 answer
331 views

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 $100$ ...
Aradhita's user avatar
1 vote
1 answer
71 views

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 $s(x)=\sigma(x)-x$. ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
71 views

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 \...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
57 views

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 ...
Jose Arnaldo Bebita Dris's user avatar
3 votes
1 answer
88 views

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 ...
C. Brendel's user avatar
1 vote
0 answers
37 views

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 $$\...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
50 views

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, ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
60 views

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 ...
user avatar
15 votes
2 answers
1k views

$\sigma(n) \equiv 1 \space \pmod{n}$ 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 ...
user avatar
1 vote
6 answers
2k views

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 ...
Skrrrrrtttt's user avatar
0 votes
1 answer
54 views

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!
Gwen Di's user avatar
  • 153
-1 votes
1 answer
75 views

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 $\...
Jose Arnaldo Bebita Dris's user avatar
0 votes
1 answer
92 views

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 $q$ ...
Jose Arnaldo Bebita Dris's user avatar
2 votes
0 answers
83 views

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 ...
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