# Prove that every odd number can be written as $(2^n-1)/A$ or $(2^n+1)/A$, $n$ & $A$ are some integers.

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?

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.

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.