thanks for helpin me. guys
1. let p be a prime, show that every prime divisor of (2^p) -1 is greater than p.
2. let n= 3^(t-1). show that 2^n = -1 (mod 3^t)
1. Suppose $\displaystyle q$ is a prime divisor of $\displaystyle 2^{p}-1$ then $\displaystyle
2^p \equiv 1\left( {\bmod .q} \right)
$ and thus, by Lagrange's Theorem, the order of $\displaystyle 2$ divides $\displaystyle
\left| {\mathbb{Z}_q ^ \times } \right|=q-1
$. (1)
Suppose $\displaystyle s>1$ is the order of $\displaystyle 2$, it cannot be 1 for otherwise $\displaystyle 2\equiv{1}(\bmod.q)$, and since $\displaystyle
2^p \equiv 1\left( {\bmod .q} \right)
$ the order must divide $\displaystyle p$, but since $\displaystyle p$ is prime this implies that $\displaystyle s=p$ that is the order of $\displaystyle 2$ is $\displaystyle p$. So, by (1), $\displaystyle p|(q-1)$ thus $\displaystyle q-1\geq{p}$ thus $\displaystyle q>p$
2. By induction, it's true for k=1, so let's assume it's true for some $\displaystyle k\in
\mathbb{Z}^ +
$ we'll show that this implies the assertion for $\displaystyle k+1$
By hypothesis $\displaystyle
2^{3^{k - 1} } \equiv - 1\left( {\bmod .3^k } \right)
$ so $\displaystyle
2^{3^{k - 1} } = 3^k \cdot s - 1
$ now: $\displaystyle
2^{3^k } = \left( {3^k \cdot s - 1} \right)^3 = 3^{k + 1} \cdot t - 1
$ ( see the terms in the binomial expansion), thus: $\displaystyle
2^{3^k } \equiv - 1\left( {\bmod .3^{k + 1} } \right)
$
In fact 2 is a primitive root module $\displaystyle 3^k$ for all $\displaystyle k\in
\mathbb{Z}^ +
$
If $\displaystyle a\equiv b(\bmod p^k) \implies a^p \equiv b^p (\bmod p^{k+1})$*.
You can prove this result by induction.
Say $\displaystyle 2^{3^{t-1}} \equiv -1 (\bmod 3^t)$ then $\displaystyle 2^{3\cdot 3^{t-1}} \equiv (-1)^3 = -1 (\bmod 3^{t+1})$ and so $\displaystyle 2^{3^t} \equiv -1(\bmod 3^{t+1})$.
*)This can be proven by binomial expanding as PaulRS did.