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Math Help - Number theory Question 2

  1. #1
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    Number theory Question 2

    Show that if p is not a Fermat prime, then there is a such that
    <br /> <br />
\left(\frac {a}{p}\right)=-1<br />

    and [a]_p is not a generator of (\mathbb Z/p)^* .

    Here <br /> <br />
\left(\frac {2}{p}\right)=-1<br />
    is legendre symbol.
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by peteryellow View Post
    Show that if p is not a Fermat prime, then there is a such that
    <br /> <br />
\left(\frac {a}{p}\right)=-1<br />

    and [a]_p is not a generator of (\mathbb Z/p)^* .
    Recall the fact that all primes of the form 2^n+1 -save from the number 2- are fermat primes -and conversly- (* that is easy to show, if you have not seen it, ask)

    Okay, first remember that there are <br />
\phi \left( {\phi \left( p \right)} \right) = \phi \left( {p - 1} \right)<br />
primitive roots ( or generators) in (\mathbb Z/p)^*

    And there are <br />
\tfrac{{p - 1}}<br />
{2}<br />
non-quadratic residues.

    So we'd like to show that <br />
\tfrac{{p - 1}}<br />
{2} > \phi \left( {p - 1} \right)<br />
occurs in our case. -remember that the set of generators is included in the set of non-quadratic residues, see here-

    We can write: <br />
p - 1 = 2^\alpha   \cdot m<br />
with <br />
\alpha;m  \in \mathbb{Z}^ +  <br />
such that <br />
\left( {m,2} \right) = 1<br />
- since p>2, the case p=2 is trivial-

    Then: <br />
\phi \left( {p - 1} \right) = \phi \left( {2^\alpha  } \right) \cdot \phi \left( m \right) = 2^{\alpha  - 1}  \cdot \phi \left( m \right) \leqslant 2^{\alpha  - 1}  \cdot m = \tfrac{{p - 1}}<br />
{2}<br />
equality occurs iff <br />
m = 1<br />

    Hence if p is not of the form 2^{\alpha}+1 (that is m>1) there's always a non-quadratic residue that is not a generator.

    But, if p is of the form 2^{\alpha}+1, then all non-quadratic residues are also generators.
    Last edited by PaulRS; January 14th 2009 at 06:03 AM.
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