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 2019-11-22, 15:32 #1 T.Rex     Feb 2004 France 2×461 Posts (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