1
$\begingroup$

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?

$\endgroup$
7
  • 2
    $\begingroup$ 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. $\endgroup$
    – kummerer94
    Jan 12 '15 at 21:49
  • $\begingroup$ 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. $\endgroup$
    – Joel
    Jan 12 '15 at 21:49
  • $\begingroup$ Look at the Wikipedia entry. $\endgroup$ Jan 12 '15 at 21:50
  • 1
    $\begingroup$ 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. $\endgroup$
    – kummerer94
    Jan 12 '15 at 21:51
  • $\begingroup$ Sorry. Editing right now. $\endgroup$ Jan 12 '15 at 21:51
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.