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2020-09-21, 07:11   #10
JeppeSN

"Jeppe"
Jan 2016
Denmark

2×34 Posts

Quote:
 Originally Posted by mathwiz http://factordb.com/index.php?query=2%5E1277-2 is fully factored. Are you saying you can therefore factor 2^1277-1?
I see an easy factorization of 2^n - 1 by induction: Suppose 2^(n-1) - 1 is factored. Then the factorization of 2^n - 2 is trivial. If we could somehow get the factorization from 2^n - 1 from that, ... /JeppeSN