20181020, 05:43  #1 
May 2004
2^{2}·79 Posts 
A tentative definition
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. 
20181020, 15:15  #2  
"Forget I exist"
Jul 2009
Dartmouth NS
20402_{8} Posts 
Quote:
10p with p>3; p 3 mod 5 at least using additive inverses. multiplicayive inverses are different. 

20181020, 19:37  #3 
"Jane Sullivan"
Jan 2011
Beckenham, UK
337 Posts 
What is tortion? Did you mean torsion?

20181020, 20:53  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{4}·7^{2}·13 Posts 
No, he meant torture. Seriously.

20181021, 01:00  #5 
Feb 2017
Nowhere
5·1,277 Posts 
Inverse (mod P)? With resect to what operation?
If "inverse" means "multiplicative inverse" we have the following:
If N is odd and the product of at least two prime factors, any two of its factors are multiplicative inverses (mod 2), and P = 2 is less than the largest prime factor of N. If N has at least three prime factors, then at least two of them are multiplicative inverses (mod 2), and P = 2 is less than the largest prime factor of N. (In the above two cases, of course, any odd factor is selfinverse (mod 2).) This leaves N = 2*p, where p is prime. If N  1 = 2*p  1 is prime, it satisfies the definition. Otherwise, it does not. In short, the numbers fitting the definition are of the form N = 2*p, where p > 2 is prime, and 2*p  1 is also prime. If you don't like the choice P = 2, you should have said so. However, even if you exclude P = 2 by fiat, your options are still limited: If N has at least four prime factors if 3 divides N, or at least three prime factors if 3 does not divide N, then N has at least one pair of factors which are multiplicative inverses (mod 3), and P = 3 is less than the largest prime factor of N. 
20181021, 05:56  #6 
May 2004
2^{2}·79 Posts 

20181021, 06:06  #7  
May 2004
474_{8} Posts 
Quote:
Am refering only to multiplicative inverses. 

20181021, 23:48  #8 
Aug 2006
2^{2}×3×499 Posts 
Up to 100, I find: 10, 15, 21, 22, 26, 30, 33, 34, 35, 39, 42, 46, 51, 55, 57, 58, 65, 66, 69, 70, 77, 78, 82, 85, 86, 87, 91, 93, 94, 95.
Up to a million there are 607926 squarefree numbers, of which 525128 meet your definition. Asymptotically the fraction is 1. Edit: I forgot to post, so Sardonicus beat me to it. 
20181022, 00:03  #9 
"Forget I exist"
Jul 2009
Dartmouth NS
20402_{8} Posts 

20181022, 04:52  #10  
May 2004
2^{2}×79 Posts 
Quote:
that a necessary condition for a squarefee composite ( with minimum 3 prime factors) to be a Devaraj number (which include Carmichael numbers) is that it should be tortion free. 

20181022, 14:43  #11 
Aug 2006
2^{2}·3·499 Posts 
A squarefree number with at least three prime factors has at least two distinct prime factors, which are multiplicative inverses mod 2.

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