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Old 2021-01-27, 09:29   #3
Nov 2014

368 Posts

Originally Posted by axn View Post
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.
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