How do I prove the $(n+1)$-th case for this equation?
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$$3^{2^{n+1}}-1 = (3^{2^{n}}-1)(3^{2^{n}}+1) = k2^{n+2}(3^{2^{n}}+1)$$
Now, note that $$3^{2^{n}}+1$$ is obviously even ($3^k$ is odd) and therefore divisble by $2$, and therefore we get: $$3^{2^{n+1}}-1 = 2^{n+2}\cdot 2m = 2^{n+3} \cdot m$$
Which shows that $$2^{n+3} \mid 3^{2^{n+1}}-1$$
Snufsan
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1The second equality should be $...=k2^{n+2}(3^{2^n}+1)$ for a certain $k\in\mathbb Z$. – idm Sep 14 '14 at 11:06
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Wow, very impressive especially considering the response time! THank you very much – Barney Chambers Sep 14 '14 at 11:09