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2021-01-27, 09:29   #3
gLauss

Nov 2014

368 Posts

Quote:
 Originally Posted by axn The residue produced is same regardless of the factors.
Ok, this was not obvious to me. I always assumed it would calculate $3^{\frac{2^n-1}{p}} \text{ mod } \frac{2^n-1}{p}$ instead (assuming p is one or more factor). Then the residues would be different.