20200212, 15:03  #1 
Mar 2018
17×31 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^{3}·3·5·7^{2} 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

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Pg primes pg(k) with k congruent to 10^n mod 41  enzocreti  enzocreti  0  20200126 19:19 
Congruent to 10^n mod 41  enzocreti  enzocreti  0  20200109 11:56 
Pg primes congruent to 1111 mod 42^2  enzocreti  enzocreti  0  20190627 12:31 
((my) mod n ) congruent to n1  smslca  Math  2  20120129 11:30 