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Old 2019-11-22, 15:32   #1
T.Rex
 
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Feb 2004
France

2×461 Posts
Default (New ?) Wagstaff/Mersenne related property

Hi,

I have no idea if this property is new. If new, I even am not sure it may be useful.
Anyway.

Let q prime >3
q=2p+1 and thus p=\frac{q-1}{2}.

Let:
N_p=2^p+1 .
M_p=2^p-1 Mersenne.
N_p M_p = 2^{2p}-1=2^{q-1}-1

Let:
W_q=\frac{2^q+1}{3} Wagstaff.

Then:
2N_pM_p+3 = 2^q-2+3 = 2^q+1 = 3W_q

Thus the property : W_q = \frac{2}{3}N_pM_p+1 . CQFD.

\alpha \mid W_q \Rightarrow \alpha = 1+2q\alpha'
thus : W_q = 1+2q\beta and 2q\beta = W_q-1 = \frac{2}{3}N_pM_p
thus : q \, \mid \, \frac{N_pM_p}{3} and thus either q \mid N_p or q \mid M_p .

Examples :
q=11 , \, p=5 , \, q \mid N_p
q=17 , \, p=8 , \, q \mid M_p
q=47 , \, p=23 , \, q \mid M_p
q=59 , \, p=29 , \, q \mid N_p
q=257 , \, p=2^7 , \, q \mid M_p
q=65537 , \, p=2^{15} , \, q \mid ?_p


Probably one should only consider cases where p is a prime or a power of 2.

If p = 2^n, then 3 divides M_p since only numbers k.2^{n+1}+1 can divide a Fermat number. Can k be 1 ?

If p is a prime, thus 3 cannot divide M_p since only numbers 1+\alpha p can divide a Mersenne number, and thus 3 divides N_p.
So, when p is a prime, when does it divide N_p and not M_p and vice-versa ??

Last fiddled with by T.Rex on 2019-11-22 at 16:24
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