(Preamble #1: In what follows, we take $\sigma=\sigma_{1}$ to be the sum of the divisors, and denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.) (Preamble #2: My sincerest apologies for the somewhat very long post — I just had to put in all the details into one place for ease of quick reference […]

Let $\mathbb{N}$ denote the set of natural numbers (i.e., positive integers). A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum of divisors. For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect. (Note that $6$ is even.) Denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$. Euler […]

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 […]

A super-perfect number is a number with $\sigma(\sigma (n))=2n$. How can I prove that every even super perfect number is from the form $n=2^k$ when $2^{k+1}-1$ is prime. I tried every way please give me some guidance

If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 \mod 9$. I know perfect numbers are of the form $(2^{p-1})(2^{p}-1)$. I have a few trials that I have done and they check out but as far as a proof I have no idea where […]

The perfect number $6$ is in the middle of the primes $5$ and $7$. It is the only perfect number with this property because odd numbers are not in the middle of two twin primes and even perfect numbers have the form $2^{n-1}(2^n-1)\ ,\ n\ge 2$ with a prime $2^n-1$, so they are greater than […]

A perfect number is the sum of its (positive) divisors (excluding itself). I am wondering if a square could be a perfect number. If it is an odd square, then, excluding itself, it has an even number of divisors which are odd. Adding them together yields an even sum. Therefore, an odd square could not […]

I’m preparing a talk on Mersenne primes, Perfect numbers and Fermat primes. In trying to provide motivation for it all I’d like to provide an application of these things. I came up with these: Applications of Mersenne numbers: signed/unsigned integers, towers of Hanoi Applications of Fermat numbers: relation to constructible polygons But for perfect numbers […]

I use the method display by Florian in [1] (in true both statments of this problem are due to Florian at 99%) to compute from $\sigma(2n)-(\sigma(n)+\sigma(n))=2^p$ (where $\sigma$ is the sum of divisor function), really there are no reason to write the second summand in the left side display as this manner, compute I said […]

How do I show that the sum of the reciprocals of divisors of a perfect number is $2$? I tried $d_i\mid n$ with $i\in\mathbb{N},\;d_i\leq n$ then $$\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+…+\frac{1}{d_i}=2$$ $$\sum_{d\mid n} \frac{1}{d}=2$$ So actually, I have to show this last equality, whereas $n$ is a perfect number.

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