I am trying to solve :
Given: $n>3$ prove that $2^n+1$ is not a power of $3$.
I just need a hint for this problem.I am trying to solve by thinking about divisibility and congruences but can't figure out the right direction.
Thanks. :)
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Andrés E. Caicedo
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Devarsh Ruparelia
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1Gerry Myerson's Answer to this slightly more general Question quotes a nice exposition of the original proof due to Gersonides in The Harmony of Numbers (1343). – hardmath Jan 11 '15 at 17:52
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Hint: In order that $3\mid(2^n+1)$, $n$ must be odd. If $m,n$ are odd numbers and $m\mid n$, then $(2^m+1)\mid(2^n+1)$. Now have a look at Zsigmondy's theorem, or prove that: $$ \nu_2(3^k-1) \leq \nu_2(k)+2.$$
Jack D'Aurizio
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Can you make it more explicit and elementary.I am not getting Zsigmondy's theorem exactly. – Devarsh Ruparelia Jan 11 '15 at 17:30
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1The point is that, as $n$ is odd an increases, $2^n+1$ tends to have newer and newer prime factors. – Jack D'Aurizio Jan 11 '15 at 17:32
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Or, by using the other inequality, consider that if $2^n = 3^k-1$, then $k$ must be big, with a size comparable to $2^n$. But that contradicts equality. – Jack D'Aurizio Jan 11 '15 at 17:34
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Hint: if $n$ is even $2^n$ is congruent to $1\bmod 3$.
If $n$ is odd you can write $2^n+1$ as $(2+1)(2^{n-1}-2^{n-2}+2^{n-3}\dots +1)$. Notice the first pair of terms is $1\bmod 3$, the second pair is also $1\bmod 3$. Therefore the part in the right is a multiple of three if and only if $n$ is a multiple of $3$.
But $2^{3k}+1=(2^k)^3+1=(2^k+1)(2^{2k}-2^k+1)$ So $2^k+1$ needs to be a power of $3$. Therefore if $k=9$ does not work we are done.
Asinomás
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Beautiful and elegant.Exactly what I was waiting for.Thanks a lot. – Devarsh Ruparelia Jan 11 '15 at 17:43