20190106, 19:41  #1 
"Angelino Desmet"
Mar 2018
Belgium
100110_{2} Posts 
Why 2^p1 and not 2^p2?
Why is 2^{p}1 special, and not 2^{p}2?

20190106, 19:43  #2 
Banned
"Luigi"
Aug 2002
Team Italia
1001010011110_{2} Posts 

20190106, 19:45  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3^{3}·13^{2} Posts 
In what sense special?
2^{p}2 is special in its own way, too 
20190106, 19:59  #4 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 

20190106, 20:02  #5 
"Angelino Desmet"
Mar 2018
Belgium
2×19 Posts 
Is that rhetorical? I really should get back into my math books.
Special as in: why are we calculating for 2^{p}1 and not 2^{p}2 as well. I guess the latter can't be prime? Last fiddled with by Blackadder on 20190106 at 20:02 
20190106, 20:17  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
no number divisible by a number other than 1 or itself is prime 2^{p}2 = 2(2^{p1}1) therefore is divisible by 2 except when p1 = 1 and therefore p=2  2^{p1}1 = 1
Last fiddled with by science_man_88 on 20190106 at 20:18 
20190106, 20:39  #7 
"Angelino Desmet"
Mar 2018
Belgium
38_{10} Posts 

20190106, 20:44  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×3^{3}×13^{2} Posts 
