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Old 2020-09-21, 07:11   #10
JeppeSN's Avatar
Jan 2016

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Originally Posted by mathwiz View Post 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
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