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Questions tagged [perfect-numbers]

Questions about or involving perfect numbers which are positive integers that are equal to the sum of their proper positive divisors.

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Can an odd perfect number be divisible by $105$?

I have a tough one today. Show that if $n$ is an odd perfect number, then not all of $3$, $5$, and $7$ are divisors of $n$. Any and all help is appreciated. Thanks very much.
12
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1answer
881 views

Can an odd perfect number be divisible by $825$?

I know that an odd perfect number cannot be divisible by $105$. I wonder if that's also the case for $825$.
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2answers
2k views

Can the cube of every perfect number be written as the sum of three cubes?

I found an amazing conjecture: the cube of every perfect number can be written as the sum of three positive cubes. The equation is $$x^3+y^3+z^3=\sigma^3$$ where $\sigma$ is a perfectnumber (well it ...
16
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2answers
686 views

If $N = q^k n^2$ is an odd perfect number and $n < q^{k+1}$, does it follow that $k > 1$?

Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. If $\sigma(M) = 2M$, then $M$ is said to be perfect. Currently, there are $49$ known examples of even perfect numbers -- on ...
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2answers
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Discussion on even and odd perfect numbers.

First of all thank you so much for answering my previous post. These are few interesting problems drawn from Prof. Gandhi lecture notes. kindly discuss: 1) If $n$ is even perfect number then $(8n +1)$...
5
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1answer
175 views

If $q^k n^2$ is an odd perfect number with Euler factor $q^k$, can $q = 73$ hold?

Call a number $N$ perfect if $\sigma(N)=2N$ where $\sigma$ is the classical sum-of-divisors function. If $N = q^k n^2$ is an odd perfect number, can $q = 73$ hold? Here is my attempt: Since $37=(...
11
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1answer
632 views

Can an odd perfect number be divisible by $165$?

I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $165$ (this is actually a stronger statement).
3
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1answer
368 views

Prove that $p^j q^i$ cannot be a perfect number for $p, q$ odd, distinct primes.

Define $\sigma(m) = \sum$ d : d|n. Prove that $p^j$$q^i$ cannot be a perfect number for $p, q$ odd, distinct primes. Attempt at Solution: I have shown that $p^k$ can never be a perfect number, and ...
17
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1answer
286 views

When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index?

Let $\sigma(x)$ denote the sum of the divisors of $x$, and denote the abundancy index of $x$ as $$I(x) = \dfrac{\sigma(x)}{x},$$ and the deficiency of $x$ as $$D(x) = 2x - \sigma(x).$$ If the equation ...
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2answers
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Applications of Perfect Numbers

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: ...
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1answer
303 views

How did Descartes come up with the spoof odd perfect number $198585576189$?

We call $n$ a spoof odd perfect number if $n$ is odd and and $n=km$ for two integers $k, m > 1$ such that $\sigma(k)(m + 1) = 2n$, where $\sigma$ is the sum-of-divisors function. In a letter to ...
2
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3answers
2k views

Sum of the reciprocals of divisors of a perfect number is $2$?

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
vote
1answer
54 views

If $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. Denote the abundancy index $I$ of $N$ as $I(N)=\...
4
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2answers
224 views

Can an odd perfect number be divisible by $5313$?

I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $5313$.
2
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1answer
124 views

Could a Mersenne prime divide an odd perfect number?

The relationship between Mersenne primes $2^r-1$ and even perfect numbers $2^{r-1}(2^r-1)$ is well-known (Euclid, Euler). In a video on the web I heard the statement that it is known that a Mersenne ...
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2answers
135 views

I attempt integrate another factor 2 in the definition of even 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 ...
0
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1answer
247 views

even perfect numbers and primes

Thank in advance to m.s.e site. I am looking for discussions/proof of the 1) If $p ($$2^r$$/2$) is an even perfect number then $p$ should be in the form of $2^r$ - 1 2) Every even perfect number ...
0
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1answer
119 views

Is $198585576189$ a member of OEIS sequence A228059?

I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me. So here it goes: Is the Descartes spoof $$\mathscr{D} = {3^2}\cdot{7^2}\...
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0answers
108 views

If we suppose, that a perfect odd number exists, can we determine whether infinite many exist?

It is currently not known whether odd perfect numbers (numbers with the property $\sigma(n)=2n$, where $\sigma(n)$ is the sum of divisors of $n$, including $1$ and $n$) exist. But suppose, a perfect ...
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3answers
3k views

Could a square be a perfect number?

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 ...
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3answers
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If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 (\bmod 9)$

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 ...
3
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1answer
85 views

Equations involving arithmetic functions, totatives and even perfect numbers

I've deduced simple relationships that satisfy each even perfecf number (even numbers $n$ for which $\sum_{d\mid n}d=2n$) and now I wondered about related conjectures. For each integer $m\geq 1$ we ...
3
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1answer
113 views

Number $N>6$, such that $N-1$ and $N+1$ are primes and $N$ divides the sum of its divisors

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 ...
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2answers
294 views

Can you prove or disprove the following list of my conjectures?

The following three statements are my own conjectures, not a homework problem. $a)$ For $n = 3, 4, 5,..$, every square integer $n^2$ can be expressed as the sum of a prime $p$ and two other primes $...
9
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0answers
289 views

A conjecture regarding odd perfect numbers

(Note: This question has now been cross-posted to MO.) Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\...
4
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0answers
148 views

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 ...
2
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2answers
153 views

When does :$\sigma(\sigma(2n))=\sigma(\sigma(n))$ and $\sigma(n)$ is sum divisors of the positive integer $n$?

Is there someone who can show me When does: $$\sigma(\sigma(2n))=\sigma(\sigma(n))$$ where : $\sigma(n)$ denotes the sum of divisors of the positive integer $n$ ? Note (1) : I came across this ...
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1answer
63 views

Can this bound for the abundancy index of $n$ be improved, given that $q^k n^2$ is an odd perfect number with $k=1$?

In what follows, set $I(x)=\sigma(x)/x$ to be the abundancy index of $x \in \mathbb{N}$, where $\sigma(x)$ is the sum of divisors of $x$. If $I(y)=2$ and $y$ is odd, then $y$ is called an odd perfect ...
0
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2answers
128 views

$n$-th term of the series 1 27 125 1000

What will be the nth term of the series 1 27 125 1000 for $n = 1$, it is 1 for $n = 2$, it is 27 for $n = 3$, it is 125 for $n = 4$, it is 1000
5
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1answer
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Super Perfect numbers

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 ...
4
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1answer
285 views

Is it possible to simplify this expression even further?

(Preamble: This question is tangentially related to this earlier one.) Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $...
3
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1answer
75 views

On the Descartes-Frenicle-Sorli conjecture and the Euler prime of odd perfect numbers

(Preamble: My apologies for the somewhat long post - I merely wanted to include all the details that I had in mind for ease of reference later.) This post is an offshoot of this earlier MSE question. ...
3
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1answer
64 views

If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, and $k=1$, does it follow that $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$?

Let $\sigma=\sigma_{1}$ be the classical sum-of-divisors function. If $N = q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q$ satisfies $q \equiv k \equiv 1 \pmod 4$), and $k=1$, does ...
3
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2answers
78 views

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

Let $\sigma(x)$ be the sum of the divisors of $x$. Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$. Here is my ...
3
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0answers
334 views

Can an odd perfect number be divisible by $101$?

Preamble - This question is an offshoot from the following earlier questions here at MSE: Can an odd perfect number be divisible by 825? Can an odd perfect number be divisible by 165? Odd perfect ...
3
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1answer
109 views

What is the common (and simplified) value for $D(q^k)D(n^2) = 2s(q^k)s(n^2)$ when $q^k n^2$ is an odd perfect number?

In an answer to an earlier question, it is shown that $$D(2^p - 1)D(2^{p-1}) = 2s(2^p - 1)s(2^{p-1}) = 2^p - 2,$$ if $2^{p-1}(2^p - 1)$ is an even perfect number, $D(x) = 2x - \sigma(x)$ is the ...
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2answers
85 views

Is $(q^k n^2 \text{ is perfect }) \iff (D(q^k)D(n^2) = 2s(q^k)s(n^2))$ only true for odd perfect numbers $q^k n^2$?

(Preamble: This question is an offshoot of this earlier MSE post.) The title says it all. Is $\bigg(q^k n^2 \text{ is perfect }\bigg) \iff \bigg(D(q^k)D(n^2) = 2s(q^k)s(n^2)\bigg)$ only true for ...
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2answers
87 views

On $\text{Lower bound}\leq \operatorname{rad}(n)$, where $n$ is an odd perfect number: reference request or what work can be done about it

For integers $n\geq 1$ we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ ...
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1answer
81 views

Do all terms $n$ of OEIS sequence A228059 have a $p$ with exponent $1$?

Do all terms $n$ of OEIS sequence A228059 have a $p$ with exponent $1$? OEIS sequence A228059: Odd numbers of the form $p^{1+4k}{r^2}$, where $p$ is prime of the form $1+4m$, $r > 1$, and $\gcd(p,...
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1answer
108 views

Is this proof regarding odd perfect numbers correct?

Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$. (That is, $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$.) Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$. ...
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0answers
72 views

Bounds for an expression involving the divisors of an odd perfect number

Let $$\sigma(x) = \sum_{l \mid x}{l}.$$ That is, let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$ (the set of natural numbers or positive integers). Set $$D(x) := 2x - \sigma(x)$$ ...
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0answers
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Conjecture: $a^3 + b^3 + c^3 = p^3 \Rightarrow x^3 + y^3 + z^3 = \big(\frac{a + b + c}{2}\big)^3$

Conjecture: Let $p$ be an even perfect number, and $a, b, c$ be positive natural numbers. There exists values for $a, b, c$ to satisfy the following equation $$a^3 + b^3 + c^3 = p^3$$ for ...
1
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1answer
146 views

What is wrong with this proof that there are no odd perfect numbers?

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 $...
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1answer
46 views

Building on work from previous MSE question 2306650 (Re: Odd Perfect Numbers)

(Note: This post builds on work from this previous MSE question.) Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an ...
0
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1answer
67 views

If $N=q^k n^2$ is an odd perfect number and $q = k$, why does this bound not imply $q > 5$?

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 ...
0
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1answer
81 views

On the quantity $\sigma(\frac{n^2 \sigma(n^2)}{D(n^2)})$ when $q n^2$ is an odd perfect number with special prime $q$

Denote the sum of the divisors of $x \in \mathbb{N}$ by $\sigma(x)$. Also, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$. If $m$ is odd and $\sigma(m)=2m$, then $m$ is called an odd perfect ...
0
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1answer
63 views

Is the Euler prime of an odd perfect number a palindrome (in base $10$), or otherwise?

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$). (That is, $2N=\sigma(N)$ where $\sigma$ is the ...
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1answer
66 views

Find all prime numbers $p$ such that $5^p+ 4p^4$ is a perfect square

Find all prime numbers $p$ such that $5^p+ 4p^4$ is a perfect square.
6
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3answers
3k views

Can powers of primes be perfect numbers?

I need to prove the following, though I'm not 100% certain I understand the definition of a perfect number. Prove that no perfect number is a power of a prime. First of all, I'm assuming that the ...
5
votes
0answers
108 views

Has it been proved that odd perfect numbers cannot be triangular?

(Note: This question has been cross-posted to MO.) Euclid and Euler proved that every even perfect number is of the form $m = \frac{{M_p}\left(M_p + 1\right)}{2}$ where $M_p = 2^p - 1$ is a prime ...