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Prove that if $2^n - 1$ is prime, then $n$ is prime for $n$ being a natural number

I've looked at https://math.stackexchange.com/a/19998

It is known that $2^n-1$ can only be prime if $n$ is prime. This is because if $jk=n$, $2^n-1=\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$ So they will only continue to alternate at twin primes. In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite

What I dont understand is how do I get:

$$\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$$

If I am looking at the wrong answer, how can I proof the above?


Then again, maybe I am indeed looking at the wrong thing?

In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite

This doesnt say why $2^{6k+2}-1$ is composite?

  • 1
    $2^{6k+2}-1=(n+1)(n-1)$ where $n=2^{3k+1}$ - it's a difference of two squares.2012-08-25
  • 0
    6k+2 and 6k+4 can't be prime because they are divisible by 2. 6k+3 is divisible by 3.2012-08-25

3 Answers 3