Why 2^p1 and not 2^p2?
Why is 2[SUP]p[/SUP]1 special, and not 2[SUP]p[/SUP]2?

[QUOTE=Blackadder;505144]Why is 2[SUP]p[/SUP]1 special, and not 2[SUP]p[/SUP]2?[/QUOTE]
Because 2[sup]p[/sup]2 is even? 
In what sense special?
2[SUP]p[/SUP]2 is special in its own way, too 
[QUOTE=Batalov;505146]In what sense special?
2[SUP]p[/SUP]2 is special in its own way, too[/QUOTE] it's twice a repunit, it shows an iteration formula for Mersenne numbers, it's 6 mod 8 for p>2 . etc. 
[QUOTE=ET_;505145]Because 2[sup]p[/sup]2 is even?[/QUOTE]
Is that rhetorical? I really should get back into my math books. [QUOTE=Batalov;505146]In what sense special? 2[SUP]p[/SUP]2 is special in its own way, too[/QUOTE] Special as in: why are we calculating for 2[sup]p[/sup]1 and not 2[sup]p[/sup]2 as well. I guess the latter can't be prime? 
[QUOTE=Blackadder;505148]Is that rhetorical? I really should get back into my math books.
Special as in: why are we calculating for 2[sup]p[/sup]1 and not 2[sup]p[/sup]2 as well. I guess the latter can't be prime?[/QUOTE] no number divisible by a number other than 1 or itself is prime 2[SUP]p[/SUP]2 = 2(2[SUP]p1[/SUP]1) therefore is divisible by 2 except when p1 = 1 and therefore p=2  2[SUP]p1[/SUP]1 = 1 
[QUOTE=science_man_88;505151]no number divisible by a number other than 1 or itself is prime. Now, 2[SUP]p[/SUP]2 = 2(2[SUP]p1[/SUP]1) therefore is divisible by 2
there is only one exception: when p1 = 1 (and p=2)  2[SUP]p1[/SUP]1 = 1 doesn't hurt primality.[/QUOTE] That makes sense. Thank you. 
[QUOTE=ET_;505145]Because 2[sup]p[/sup]2 is even?[/QUOTE]
Just to state the obvious, what Luigi 'ET_' said: the peoplespeak word 'even' means (in mathspeak) 'divisible by 2'. No more, no less. If something is even  it cannot be prime, except if something is '2' itself. 
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