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2018-10-23, 04:54   #12

May 2004

22·79 Posts

Quote:
 Originally Posted by devarajkandadai Let N be a squarefree composite number with r factors, p_1,...p_r. Then we can define N as a tortionfree number if atleast two of its factors are inverses mod(P),where P is a prime number less than the largest prime factor of N.
We can now go to the next level. Note that 3 and 17 ( which are prime factors of the Carmichael number 561) are not only inverses (mod 5) but also inverses (mod 5^2).
(to be continued).

2018-10-24, 04:47   #13

May 2004

13C16 Posts
A tentative definition

Quote:
 Originally Posted by devarajkandadai We can now go to the next level. Note that 3 and 17 ( which are prime factors of the Carmichael number 561) are not only inverses (mod 5) but also inverses (mod 5^2). (to be continued).
We can now define tortion free composite numbers as follows:
Let N be a squarefree composite number such that atleast two of its prime factors are inverses (mod P^k) where k is a natural number. Then N is a tortion free number of degree k. (P is a prime number less than the largest prime factor of N).

Last fiddled with by devarajkandadai on 2018-10-24 at 04:51 Reason: To make it clearer

2018-10-24, 05:51   #14

May 2004

22×79 Posts
A tentative definition

Quote:
 Originally Posted by devarajkandadai We can now define tortion free composite numbers as follows: Let N be a squarefree composite number such that atleast two of its prime factors are inverses (mod P^k) where k is a natural number. Then N is a tortion free number of degree k. (P is a prime number less than the largest prime factor of N).
Example: 11305(ref a 104017) is tortion free of degree 2.

2018-10-24, 11:04   #15

May 2004

22×79 Posts

Quote:
 Originally Posted by devarajkandadai We can now define tortion free composite numbers as follows: Let N be a squarefree composite number such that atleast two of its prime factors are inverses (mod P^k) where k is a natural number. Then N is a tortion free number of degree k. (P is a prime number less than the largest prime factor of N).
This class of composite numbers will be called "Madhavan numbers" .

2018-10-24, 12:26   #16
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

841810 Posts

Quote:
 Originally Posted by devarajkandadai This class of composite numbers will be called "Madhavan numbers" .
wouldn't powers of this form allow for Beal's conjecture solutions ?

2018-10-24, 13:45   #17
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by science_man_88 wouldn't powers of this form allow for Beal's conjecture solutions ?
sorry still thinking in the wrong inverses

2018-10-24, 20:34   #18
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 Originally Posted by devarajkandadai We can now define tortion free composite numbers as follows: Let N be a squarefree composite number such that atleast two of its prime factors are inverses (mod P^k) where k is a natural number. Then N is a tortion free number of degree k. (P is a prime number less than the largest prime factor of N).
I have an example with degree k = 4031399, can anyone do better?

2018-10-28, 05:39   #19

May 2004

22·79 Posts
A tentative definition

Quote:
 Originally Posted by devarajkandadai This class of composite numbers will be called "Madhavan numbers" .
Time to define inverses of higher order: let x and y be such that xy+1 = a*p^k+1 where a is a constant belonging to N, k is a natural number and p is a prime number. Then x and y are inverses of degree k.

2018-10-29, 12:33   #20

May 2004

13C16 Posts
A tentative definition

Quote:
 Originally Posted by CRGreathouse I have an example with degree k = 4031399, can anyone do better?
Charles, pl give details of above number.

 2018-10-29, 14:41 #21 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 89×113 Posts Hint: P=2, and P^k+1 is a semiprime.
2018-10-29, 15:13   #22
Dr Sardonicus

Feb 2017
Nowhere

2·11·283 Posts

Quote:
 Originally Posted by CRGreathouse I have an example with degree k = 4031399, can anyone do better?
While rummaging around on line, I blundered into the candidates k = 13347311 and k = 13372531 due to R. Propper (Sep. 2013).

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