[seqfan] Re: " Little Fermat theorem" for twin primes

Max Alekseyev maxale at gmail.com
Wed Oct 12 23:28:33 CEST 2011

Equivalence of (1) and (2) is a simple consequence of Chinese
Remainder Theorem (for an odd n and b coprime to n*(n+1)).
First notice that for an odd n,
n * (-1/2) == 1 (mod n+2)
(n+2) * (1/2) == 1 (mod n).
Second, the system (1) can be rewritten as
b^(n+1) == b^2 (mod n)
b^(n+1) == 1 (mod n+2).
By CRT this system is equivalent to
b^(n+1) == b^2*(n+2)/2 + n * (-1/2)   (mod n*(n+2))
which is just another form of (2).


On Wed, Oct 12, 2011 at 8:39 AM, Vladimir Shevelev <shevelev at bgu.ac.il> wrote:
> Dear SeqFans,
> Trivially " Little Fermat theorem" for twin primes {n,n+2} sounds as the following:  for every b prime to n*(n+2),
>                      b^(n-1)==1(mod n) and b^(n+1)==1 mod(n+2).          (1)
> The system of these two congruences is not equivalent to (b^(n-1)-1)*(b^(n+1)-1)==0 mod(n*(n+2)) (a simple counterexample n=7, since b^2==1(mod 3)).  However, in Little Fermat theorem there is only one congruence. I found another simple congruence
>                      2*(b^(n+1)-1)==(b^2-1)*(n+2) (mod n*(n+2)).          (2)
>  Now n=7 is not a solution, e.g., for b=2. A question: Is it true that (2) yields (1)?
> If it is true, then the "Carmichael pseudo twin primes"  can be characterized by (2) only; otherwise, there could be exist other "Carmichael pseudo twin primes" satisfying (2) for all  b prime to n*(n+2).
> Note that the sequence of Carmichael pseudo twin primes (I submitted lesser of them in A194231), i.e. numbers {n,n+2}satisfying (1) for every  b prime to n*(n+2) such that from them there exists at least one composite number,  starts with 561, 1103, 2465, 2819, 6599, 29339,...
> Regards,
> Vladimir
>  Shevelev Vladimir‎
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