Timeline for Another Question About Powers mod p
Current License: CC BY-SA 3.0
3 events
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| Aug 14, 2011 at 23:45 | comment | added | Gerhard Paseman | More generally, for n=hk composite, one can "remove" some factors of n from consideration, reducing the problem to $(h^h)^k = ((a^a)(b^b))^k$ mod $p$, or not equals mod $p$. For $h=2$, this gives that $p$ must be a factor of $4^k - 1$. Gerhard "Ask Me About System Design" Paseman, 2011.08.14 | |
| Jul 31, 2011 at 22:41 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
added 18 characters in body
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| Jul 31, 2011 at 22:27 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |