So, I'm wondering if, given a Mersenne prime, $2^p-1$ whether or not $p$ also divides $2^p-2$. From the wikipedia page on Mersenne prime, it seems like it. For example, $7 = 2^3-1$ is a Mersenne prime and 6 is divisible by 3. Similarly, $8191 = 2^{13}$ and 8190 is divisible by 13.
Asked
Active
Viewed 127 times
0
-
2http://en.wikipedia.org/wiki/Fermat's_little_theorem says $n^p-n\equiv0\pmod p$ for any prime $p$ which is necessary for $2^p-1$ to be prime. See http://math.stackexchange.com/questions/186587/prove-that-if-2n-1-is-prime-then-n-is-prime-for-n-being-a-natural-numb – lab bhattacharjee Nov 12 '13 at 07:57
1 Answers
1
Since any prime $p$ with $\gcd(2,p)=1$ divides $2^{p-1}-1$ (Little Fermat, Euler) it does also divide $ 2^p-2 = 2(2^{p-1}-1)$ . The reference to the property of Mersenne-primes is a hidden hint: this is the remark, which defines $p$ being element of the set of primes: $2^p-1$ being a Mersenneprime implies $ p \in \mathbb P$ so we are talking about a prime $p$ here ...
Gottfried Helms
- 34,920
-
-
@lab: :-) I needed a time to get the catch... true; I did not include that; I seem to be too much used to the convention $p$ meaning a prime by default that I forget asking. Thanks for pointing at this! – Gottfried Helms Nov 12 '13 at 08:56