Timeline for Another Question About Powers mod p
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31, 2011 at 22:27 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Jul 28, 2011 at 22:50 | comment | added | Gerhard Paseman | n=3. In rewriting the problem n=2 should also be excluded. Gerhard "Ask Me About System Design" Paseman, 2011.07.28 | |
| Jul 28, 2011 at 22:33 | comment | added | Daniel Litt | @Gerhard: What is the counterexample? | |
| Jul 28, 2011 at 20:59 | comment | added | Gerhard Paseman | Actually, p=23 has a counterexample, if I read the question right. Gerhard "Ask Me About Small Cases" Paseman, 2011.07.28 | |
| Jul 28, 2011 at 19:43 | comment | added | Greg Martin | I'm not sure this is helpful, but when $a+b=n$ and $a$ and $n$ are relatively prime, then $n^n - a^a b^b$ is the discriminant of the trinomial $x^n - x^a + 1$ (equivalently, of $x^n - x^b + 1$). | |
| Jul 28, 2011 at 19:35 | comment | added | Greg Martin | Any specific choice of $a$ and $b$ works for all but finitely many primes $p$.... | |
| Jul 28, 2011 at 16:38 | comment | added | Gerhard Paseman | Let n be even and a=b. This works for all but finitely many primes p. Perhaps a simple variation can handle the remaining p, and you might do something similar for composite n. Gerhard "Ask Me About System Design" Paseman, 2011.07.28 | |
| Jul 28, 2011 at 14:12 | history | asked | user4535 | CC BY-SA 3.0 |