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I found an interesting fact that every odd number can be written as

$(2^n-1)/A$ or $(2^n+1)/A$, where $n$ & $A$ are some integers.

If the odd number is $N$, then $n ≤ (N-1)/2$.

I have checked from $3$ to $101$ and it is true for all these odd numbers.

ex. $101=(2^{50}+1)/11147523830125$

Is there a general proof for this odd number expression form?

Or a proof that this statement is wrong?

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For all odd numbers $N$, $2^{\phi(N)} \equiv 1 \pmod{N}$.

$\phi(N)$ is even. Hence $2^ { \frac{\phi(N)}{2}} \equiv \pm 1 \pmod{N}$

Now show that $ \frac{ \phi(N)}{2} \leq \frac{ N-1}{2}$. Equality holds when $N$ is a prime.

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N is any odd number, then N and 2 are co-prime. From Euler’s Theorem:

$2^{\varphi (N)}\equiv 1 \pmod{N}$

$\varphi (N)$ is Euler’s totient function, $\varphi (N) \leq N-1$

$2^{\varphi (N)}-1\equiv 0 \pmod{N}$

$\varphi (N)$ is even for $N \geq 3$

$(2^{\frac{\varphi (N)}2}-1)(2^{\frac{\varphi (N)}2}+1)\equiv 0 \pmod{N}$

$2^{\frac{\varphi (N)}2}-1\equiv 0 \pmod{N}\quad or\quad2^{\frac{\varphi (N)}2}+1\equiv 0 \pmod{N}$

$N=\frac{2^{\frac{\varphi (N)}2}-1}A\quad or\quad N=\frac{2^{\frac{\varphi (N)}2}+1}A,\quad \frac{\varphi (N)}2\leq \frac{(N-1)}{2}$

where A is some positive integer.

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