$ n|2^{n-1}+1$ if and only if n=1.

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Quote: Let $ n$ be a natural number, and $ n = 2^{2007}k + 1$, such that $ k$ is an odd number. Prove that \[ n\not|2^{n - 1} + 1\] Assume that $ n\mid 2^{n - 1} + 1 = \left (2^k \right )^{2^{2007}} + 1$. Let $ p$ be an arbitrary prime divisor of $ n$ (Of course, $ p$ must not less than $ 3$) and set $ 2^k = t$. We have $ t^{2^{2007}}\equiv - 1 \pmod {p}$ $ \Longrightarrow$ $ t^{2^{2008}}\equiv 1 \pmod {p}$. Let $ d = \text {ord}_{p}(t)$. It is followed that $ d\mid 2^{2008}$ $ \Longrightarrow$ $ d = 2^{2008}$. Thus $ 2^{2008} = d\mid p - 1$, i.e. $ p\equiv 1 \pmod {2^{2008}}$. Therefore: \[ n\equiv 1 \pmod {2^{2008}} \Longrightarrow 2^{2007}k\equiv 0 \pmod {2^{2008}}\Longrightarrow 2\mid k.\] wich leads to the contradiction. The result of the problem is lead as follow.

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