[seqfan] Re: Next conjecture help needed!
Max Alekseyev
maxale at gmail.com
Sat Nov 22 20:25:06 CET 2008
Artur,
In the original message on this topic I proved the following statement:
if p is not a Mersenne prime then there exists another prime q such that
p and q together divide or not divide any Mersenne number 2^n - 1.
That's exactly your conjecure if I got it correctly.
Regards,
Max
On Sat, Nov 22, 2008 at 11:21 AM, Artur <grafix at csl.pl> wrote:
> Dear Max,
> I was understand now, you only proove that doesn't exist such q when p is
> Mersenne prime and nothing else.
> But of course any Mersenne prime don't divide any composite number which
> divisors are candidates to my conjecture.
> That we are in start point.
> Best wishes
> Artur
>
>
> Max Alekseyev pisze:
>>
>> On Sat, Nov 22, 2008 at 10:53 AM, Artur <grafix at csl.pl> wrote:
>>
>>>
>>> Dear Max,
>>> 1)
>>>
>>>>
>>>> I believe Mersenne primes
>>>>
>>>
>>> I hope your "Mersenne primes" you mean Mersenne numbers.
>>> 2)
>>>
>>>
>>>>
>>>> an odd prime number p there exists a
>>>> smallest positive integer m (called the multiplicative order of 2
>>>>
>>>
>>> <modulo p) such that p divides 2^m - 1.
>>>
>>> For every odd prime number p there exists a smallest positive integer m
>>> (called the multiplicative order of 2 modulo p) such that p divides 2^m -
>>> 1.
>>>
>>> but this not say that can't existed such q that also divided 2^m - 1 for
>>> that same value m (p and q have that same multiplicative order of 2
>>> modulo p
>>> and q)
>>>
>>
>> 1) Saying "Mersenne prime" I meant Mersenne prime, not just number - see
>> below:
>>
>> 2) Such q does not exist if p is a Mersenne prime, in which case p =
>> 2^m - 1 and there is no any non-trivial co-factor of p in the
>> factorization of 2^m - 1.
>> For example, let p = 2^2 - 1 = 3 (a Mersenne prime).
>> Then p does not have any "mate" since, for example, 3 is the only
>> common factor of 2^4-1=3*5 and 2^6-1=63=3*3*7.
>>
>> Regards,
>> Max
>>
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>>
>
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