20160908, 04:02  #23  
"Rashid Naimi"
Oct 2015
Remote to Here/There
100100011101_{2} Posts 
Quote:
4^n64= 64(4^(n3)1) Dropping multiple of power of 2: >>4^m1= 2^(2q)1 The function has 1/2 the number of infinite instances of 2^n1, which is still infinite. Is that enough to guarantee divisibility by all primes? Yes, I think so since 2^(2q)1 will always be divisible by 2^q1. The 2 functions are equivalent in this regard. 

20160908, 07:11  #24 
Aug 2006
5,987 Posts 
You could just as easily take 15^n  1 or 6^n  36.

20160908, 11:45  #25 
"Forget I exist"
Jul 2009
Dartmouth NS
10000011100010_{2} Posts 

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