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From: recoder on 3 May 2010 08:27
Dear All,
Sometimes I see this expression:
Let Mq=u*v = (2aq+1)(2bq+1) where u and v are primes and Mq a mersenne
number.
This is true if there are only 2 prime factors of Mq.
If we have more than 2 factors then (2bq+1) is not a prime but a
product of primes. then (2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) ....

Is there a way to detect if Mq has only 2 factors without trial
factoring ?

In addition,
Is there a known rule for (2bq+1) as
(2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) .... is not a prime and all
the prime divisors are congruent to +- 1 mod 8.
From: Pubkeybreaker on 3 May 2010 10:06
On May 3, 8:27 am, recoder <kurtulmeh...(a)gmail.com> wrote:
> Dear All,
> Sometimes I see this expression:
> Let Mq=u*v = (2aq+1)(2bq+1) where u and v are primes and Mq a mersenne
> number.
> This is true if there are only 2 prime factors of Mq.
> If we have more than 2 factors then (2bq+1) is not a prime but a
> product of primes. then (2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) ....
>
> Is there a way to detect if Mq has only 2 factors without trial
> factoring ?
>
> In addition,
> Is there a known rule for (2bq+1) as
> (2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) .... is not a prime and all
> the prime divisors are congruent to +- 1 mod 8.

You already asked this in the Mersenne Forum. It has already been
answered.
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