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For $2^n-1$, where $n$ is a prime number, is it true that you don't always get a Mersenne prime? Remember, a Mersenne prime is a number that has a power of two subtracted by one and is then prime:$$n|2^n-1$$$$2|2^2-1$$$$=3$$$$3|2^3-1$$$$=7$$$$5|2^5-1$$$$=31$$$$7|2^7-1$$$$=127$$$$11|2^{11}-1$$$$=2,047$$ Well, $2,047=23\times89$
So, 11 can't be used for $n$, even though it's a prime number. How does this happen that $n$ can't always be a prime number to fit into $2^n-1$ to give out a Mersenne prime?

Peter
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ReliableMathBoy
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    No proof but reasonable: If you're hypothesis would hold, we could calculate endlessly many Mersenne primes just by taking the last calculated prime $p$ and using it to create $2^p - 1$.

    Since people try hard to find these primes, you could follow that your hypothesis won't stand.

     – kummerer94 Jan 12 '15 at 21:49
  • What we know is that often when we plug a prime into the equation $f(n)=2^n-1$ we find another prime. It is unknown whether there are even an infinite number of such primes. – Joel Jan 12 '15 at 21:49
  • Look at the Wikipedia entry. – Rory Daulton Jan 12 '15 at 21:50
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    The inclusion holds the other way around. If we have got a Mersenne prime $2^p - 1$ we can conclude that $p$ is a prime. – kummerer94 Jan 12 '15 at 21:51
  • Sorry. Editing right now. – ReliableMathBoy Jan 12 '15 at 21:51
  • There are many primes upto the $p$ giving the largest known (Mersenne) prime, but very few known Mersenne primes. – Peter Jan 12 '15 at 21:54
  • So, the chance of getting a prime is small even if $p$ is prime, and $0$ if $p$ is not a prime. – Peter Jan 12 '15 at 21:55

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This is a bit challenging to answer. In short, there is no reason why $2^p - 1$ would always give a prime number. You have just given evidence to this fact.

A Mersenne prime is a prime of the form $2^p - 1$. This definition does not mean that every number of the form $2^p - 1$ is prime.

davidlowryduda
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