Main Question What is wrong with this proof that there are no odd perfect numbers? The “Proof” Euler proved that an odd perfect number $N$, if any exists, must take the form $N = q^k n^2$ where $q$ is the Euler prime satisfying $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$. Denote the sum […]

Let $q$ be an odd composite integer and $\sigma(q)$ the sum of the positive divisors of $q$. For what $q$ is it true that $$(\sigma(q)-q) \mid (q-1) \;?$$ If $q$ is prime, it is clear that it is true. Is it only true for all primes $q$? Thanks for the help.

Prove that if $k$ is highly abundant or highly composite and $q,p$ are the nearest primes with $q+1<k<p-1$, then $k-q,p-k$ are primes. This immediately implies that all highly abundant and highly composite numbers are the sum of two primes and the difference of two primes. I’ve checked this for the first $1231$ highly abundant numbers […]

(Note: This question has been cross-posted to MO.) Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. For example, $\sigma(6)=1+2+3+6=12$ and $\sigma(28)=1+2+4+7+14+28=56$. Denote the abundancy index $I$ of $x$ by $$I(x)=\frac{\sigma(x)}{x}.$$ If a positive integer $y$ is one of at least two solutions of $$I(y)=\frac{a}{b}$$ for a given rational number $a/b$, […]

Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$. where $\sigma (n)$ is the sum of all the divisors of $n$ and $\sum\nolimits_{d|n} f(d)$ is the summation of $f$ at each $d$ where $d$ is the divisor of $n$. I have written $n=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}…….p_k^{\alpha_k}$ then:- $$\begin {align*} \sum\nolimits_{d|n} \frac{1}{d}&=\frac{d_2.d_3……d_m+d_1.d_3……d_m+……..+d_1.d_2.d_3……d_{m-1}}{d_1.d_2.d_3……d_m} \\&=\frac{d_2.d_3……d_m+d_1.d_3……d_m+……..+d_1.d_2.d_3……d_{m-1}}{p_1^{1+2+…+\alpha_1}p_2^{1+2+…+\alpha_2}p_3^{1+2+….+\alpha_3}…….p_k^{1+2+….+\alpha_k}} \\ \end{align*}$$ where […]

Let $\sigma=\sigma_{1}$ denote the classical sum of divisors. For instance, $\sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28$. Let $x \in \mathbb{N}$ ($\mathbb{N}$ is the set of natural numbers/positive integers). Denote the number $2x – \sigma(x)$ by $D(x)$. We call $D(x)$ the deficiency of $x$. Now, let $m, […]

I managed to find an interesting inequality containing the sum of the divisors of a number and the number of them, using AM-GM between each 2 of the divisors of it: $\sigma(n)\geq\sqrt n.d(n)+(\sqrt n-1)^2$. I tried to use this to find a bound for the average of divisors of $n$, and I got: $\frac{\sigma(n)}{d(n)}\geq\sqrt n+\frac{n-2\sqrt […]

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the deficiency of the deficient number $x>1$ by $D(x)=2x-\sigma(x)$. Since $x>1$ is deficient, we have $D(x) \geq 1$. (Note that $D(x) \neq 0$ since deficient numbers cannot be perfect.) Additionally, since $\sigma(x) \geq x+1$ for all $x>1$, it follows that $D(x)=2x-\sigma(x) \leq x-1$. Thus, we have the […]

This is an exercise from Apostol’s number theory book. How does, one prove that $$ \frac{\sigma(n)}{n} < \frac{n}{\varphi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n} \quad \text{if} \ n \geq 2$$ I thought of using the formula $$\frac{\varphi(n)}{n} = \prod\limits_{p \mid n} \Bigl(1 – \frac{1}{p}\Bigr)$$ but couldn’t get anything further. Notations: $\sigma(n)$ stands for the sum of divisors $\varphi(n)$ […]

Let $\sigma$ be the divisor sum function, $\gamma$ the Euler-Mascheroni constant and $n>5040$. Robin showed that if the inequality$$\displaystyle \sigma(n)<e^{\gamma}n\log\log n$$ ever fails, it does infinitely often. It is quite intuitive that infinitely many of the counterexamples (if not all) would be superabundant, i.e. natural numbers $a$ such that $\displaystyle \frac{\sigma(a)}{a}>\frac{\sigma(b)}{b}$ for all $b<a$. My […]

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