20200212, 15:03  #1 
Mar 2018
1017_{8} Posts 
N congruent to 2^2^n mod(2^2^n+1)
92020 is congruent to 2^(2^2) mod (2^(2^2)+1) where 2^(2^2)+1 is a Fermat prime
Are there infinitely many numbers N congruent to (2^(2^n)) mod (2^(2n)+1) where (2^(2n)+1) is a Fermat prime? 
20200212, 15:13  #2  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×1,531 Posts 
Quote:
16 mod 17 = 16 33 mod 17 = 16 50 mod 17 = 16 ... So what. Last fiddled with by retina on 20200212 at 15:14 

20200212, 15:14  #3 
Mar 2018
17·31 Posts 
ok
ok nevermind

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