I want to start with a contradiction assuming $a$ isn't $2$ but no idea how to do that so I'm assuming it's wrong. Any help?
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1Hint: the polynomial $x^n-1$ is divisible by $x-1$. – lulu Jan 09 '19 at 18:06
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Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$. – fleablood Jan 09 '19 at 18:22
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Hint: Use the difference of $n^{th}$ powers:
$$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}\right)$$
Apply this to $a^b-1$:
$$\implies a^b-1 = \color{blue}{(a-1)}\left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1\right)$$
What happens if $\color{blue}{a > 2}$?
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