Question

On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part IV

score 278 · Answer 1 · 2026-03-27 15:18:58

278

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On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part IV

Published on 27 Mar 2026 - 15:18

score 140 · Answer 2 · 2026-03-27 15:19:03

140

Views

On a possible proof for $p^k < m$, if $p^k m^2$ is an odd perfect number with special prime $p$

Published on 27 Mar 2026 - 15:19

score 59 · Answer 3 · 2026-03-27 15:19:00

59

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On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part V

Published on 27 Mar 2026 - 15:19

score 158 · Answer 4 · 2021-10-03 12:55:27

158

Views

Proof verification - Discarding a subcase of odd perfect numbers

Published on 03 Oct 2021 - 12:55

score 346 · Answer 5 · 2026-02-23 12:08:47

346

Views

Show that if $n$ is a positive integer greater than $1$, then the Mersenne number $M_n$ cannot be the power of a positive integer.

Published on 23 Feb 2026 - 12:08

score 139 · Answer 6 · 2026-03-27 15:18:58

139

Views

On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part VI

Published on 27 Mar 2026 - 15:18

score 77 · Answer 7 · 2021-10-08 04:11:30

77

Views

Will it be possible to compute a factored expression for $n^2 - q^k$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Published on 08 Oct 2021 - 4:11

score 95 · Answer 8 · 2021-10-12 10:24:46

95

Views

On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$

Published on 12 Oct 2021 - 10:24

score 177 · Answer 9 · 2021-10-19 07:04:32

177

Views

On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{\frac{k-1}{2}} n^2$ - Part III

Published on 19 Oct 2021 - 7:04

score 66 · Answer 10 · 2021-11-28 05:27:48

66

Views

Is it possible to improve on these bounds for $\frac{\varphi(n)}{n}$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Published on 28 Nov 2021 - 5:27

score 1.2k · Answer 11 · 2021-12-03 05:39:05

1.2k

Views

Elementary Proof of No Odd Perfect Numbers

Published on 03 Dec 2021 - 5:39

score 118 · Answer 12 · 2021-12-08 05:03:43

118

Views

On the inequality $I(q^k)+I(n^2) \leq \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}$ where $q^k n^2$ is an odd perfect number

Published on 08 Dec 2021 - 5:03

score 105 · Answer 13 · 2021-12-17 05:06:41

105

Views

If $k=1$, then the index of an odd perfect number $p^k n^2$ (as defined by Chen and Chen (2014)) is not squarefree.

Published on 17 Dec 2021 - 5:06

score 155 · Answer 14 · 2021-12-19 22:33:59

155

Views

Help with "A Simpler Dense Proof regarding the Abundancy Index."

Published on 19 Dec 2021 - 22:33

score 160 · Answer 15 · 2021-12-25 05:16:06

160

Views

On odd perfect numbers and a GCD - Part V

Published on 25 Dec 2021 - 5:16

List Question

On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part IV

On a possible proof for $p^k < m$, if $p^k m^2$ is an odd perfect number with special prime $p$

On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part V

Proof verification - Discarding a subcase of odd perfect numbers

Show that if $n$ is a positive integer greater than $1$, then the Mersenne number $M_n$ cannot be the power of a positive integer.

On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part VI

Will it be possible to compute a factored expression for $n^2 - q^k$, if $q^k n^2$ is an odd perfect number with special prime $q$?

On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$

On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{\frac{k-1}{2}} n^2$ - Part III

Is it possible to improve on these bounds for $\frac{\varphi(n)}{n}$, if $q^k n^2$ is an odd perfect number with special prime $q$?

Elementary Proof of No Odd Perfect Numbers

On the inequality $I(q^k)+I(n^2) \leq \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}$ where $q^k n^2$ is an odd perfect number

If $k=1$, then the index of an odd perfect number $p^k n^2$ (as defined by Chen and Chen (2014)) is not squarefree.

Help with "A Simpler Dense Proof regarding the Abundancy Index."

On odd perfect numbers and a GCD - Part V

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