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Thread: 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 $\displaystyle
    f(x) = x^2 + 1
    $

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

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

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

    b) This one is easier. Suppose there were finitely many $\displaystyle
    \rho _1 ,...,\rho _n
    $, and consider $\displaystyle
    4 \cdot \left( {\rho _1 \cdot ... \cdot \rho _n } \right) + 3
    $
    Last edited by PaulRS; Apr 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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