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Math Help - Infintely many primes of the form:

  1. #1
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    Infintely many primes of the form:

    Hi,

    Prove that there exists infinetly many primes of the form:

    a) 4q + 1 for some q in N U {0}
    b) 4q + 3 for some q in N U {0}

    Any help would be greatly appreciated.
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  2. #2
    Super Member PaulRS's Avatar
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    a) consider the polynomial <br />
f(x) = x^2  + 1<br />

    Fix x and suppose <br />
  {f\left( x \right)} <br />
has a prime divisor p>2 then <br />
x^2  \equiv  - 1\left( {\bmod .p} \right)<br />

    By Fermat's Little Theorem: 1\equiv{x^{p-1}}=<br />
(x^2)^{\tfrac{p-1}{2}}  \equiv (-1)^{\tfrac{p-1}{2}}\left( {\bmod .p} \right)<br />
and so <br />
\left. 4 \right|\left( {p - 1} \right)<br />
 that is  <br />
p \equiv 1\left( {\bmod .4} \right)<br />
(1)

    Now assume there were finitely many primes \equiv 1 (\bmod.4) let them be <br />
\theta _1 ,...,\theta _n <br />
and consider f(<br />
2\cdot \theta _1  \cdot ... \cdot \theta _n )<br />
you can check that <br />
f\left( {2\cdot \theta _1  \cdot ... \cdot \theta _n } \right) \equiv 1\left( {\bmod .\theta _i } \right)<br />
<br />
\forall i \in \left\{ {1,...,n} \right\}<br />
and so it is not divisible by a prime \equiv{1}(\bmod.4) however this contradicts (1), since f(2\cdot<br />
\theta _1  \cdot ... \cdot \theta _n )<br />
must have at least one odd prime divisor (since it's odd>1). \square

    b) This one is easier. Suppose there were finitely many <br />
\rho _1 ,...,\rho _n <br />
, and consider <br />
4 \cdot \left( {\rho _1  \cdot ... \cdot \rho _n } \right) + 3<br />
    Last edited by PaulRS; April 14th 2009 at 03:59 PM. Reason: Typo
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  3. #3
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    Thanks alot.

    I understand.
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