Show that if $\displaystyle 2^{p} -1$ is prime, then $\displaystyle p$ must be prime
I recommend proving the "contrapositive". Assume p is NOT prime and show that $\displaystyle 2^p- 1$ is not prime.
It will help to use $\displaystyle x^n- 1= (x- 1)(x^{n-1}+ x^{n-2}+ x^{n-3}+ \cdot\cdot\cdot+ x+ 1)$. You can't apply that to $\displaystyle 2^p- 1$ directly because, taking x= 2, x- 1= 2- 1= 1 so that's not a new factor.
Another related question, prove that 2^2k - 1 is divisible by 3 always.
Salahuddin
Maths online