prove that n is composite, then 2^(n)-1 is composite. I need a prove, but I just want to know can I also prove using a counter example.
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Originally Posted by Sally_Math prove that n is composite, then 2^(n)-1 is composite. I need a prove, but I just want to know can I also prove using a counter example. Hint: $\displaystyle x^m - 1 = (x-1)(x^{n-1}+x^{n-2}+...+x+1)$. Now assume that $\displaystyle n=ab$ and consider $\displaystyle \left( 2^a\right)^b - 1$ and expand.
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