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I attempted this by induction: Here is what I did

For $k=1, 2^{3^{1}}+1 = 2^{3}+1 = 9 = 3^{2}$ is divisible by $3^{1}$ , so the result is true for $k=1$

Now I assume the result to be true for $k=m$,

$2^{3^{m}}+1$ is divisible by $3^m$. To show the result to be true for $k=(m+1)$,

$2^{3^{m+1}}+1 = 2^{3^m} \times 2^3+1$ and I was stuck here.

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    Do you mean $2^{3^{(m+1)}}+1$ in the last step2012-03-14

5 Answers 5