Show that if $\displaystyle 2^{p} -1$ is prime, then $\displaystyle p$ must be prime

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- Oct 18th 2012, 09:02 AMmaximus101Number theory question 2-3
Show that if $\displaystyle 2^{p} -1$ is prime, then $\displaystyle p$ must be prime

- Oct 18th 2012, 09:12 AMHallsofIvyRe: Number theory question 2-3
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. - Oct 26th 2012, 02:51 AMSalahuddin559Re: Number theory question 2-3
Another related question, prove that 2^2k - 1 is divisible by 3 always.

Salahuddin

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