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Math Help - number theory

  1. #1
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    number theory

    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)
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  2. #2
    Super Member PaulRS's Avatar
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    1. Suppose q is a prime divisor of 2^{p}-1 then <br />
2^p  \equiv 1\left( {\bmod .q} \right)<br />
and thus, by Lagrange's Theorem, the order of 2 divides <br />
\left| {\mathbb{Z}_q ^ \times  } \right|=q-1<br />
. (1)

    Suppose s>1 is the order of 2, it cannot be 1 for otherwise 2\equiv{1}(\bmod.q), and since <br />
2^p  \equiv 1\left( {\bmod .q} \right)<br />
the order must divide p, but since p is prime this implies that s=p that is the order of 2 is p. So, by (1), p|(q-1) thus q-1\geq{p} thus q>p

    2. By induction, it's true for k=1, so let's assume it's true for some k\in<br />
\mathbb{Z}^ +  <br /> <br />
we'll show that this implies the assertion for k+1

    By hypothesis <br />
2^{3^{k - 1} }  \equiv  - 1\left( {\bmod .3^k } \right)<br />
so <br />
2^{3^{k - 1} }  = 3^k  \cdot s - 1<br />
now: <br />
2^{3^k }  = \left( {3^k  \cdot s - 1} \right)^3  = 3^{k + 1}  \cdot t - 1<br />
( see the terms in the binomial expansion), thus: <br />
2^{3^k }  \equiv  - 1\left( {\bmod .3^{k + 1} } \right)<br />

    In fact 2 is a primitive root module 3^k for all k\in<br />
\mathbb{Z}^ +  <br />
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  3. #3
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    Quote Originally Posted by felixmcgrady View Post
    2. let n= 3^(t-1). show that 2^n = -1 (mod 3^t)
    If a\equiv b(\bmod p^k) \implies a^p \equiv b^p (\bmod p^{k+1})*.

    You can prove this result by induction.
    Say 2^{3^{t-1}} \equiv -1 (\bmod 3^t) then 2^{3\cdot 3^{t-1}} \equiv (-1)^3 = -1 (\bmod 3^{t+1}) and so 2^{3^t} \equiv -1(\bmod 3^{t+1}).

    *)This can be proven by binomial expanding as PaulRS did.
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