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Re: Integer iff prime

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  • Maximilian Hasler
    ... That s what I meant to say. And p^2+2^2-9 = p^2-5 (p-1)^2 for 2p-1 5 p 3 so this can t be a square, either. Which only leaves the trivial case q=3
    Message 1 of 6 , Apr 25, 2009
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      > If p,q>3 are prime then p^2=1(mod 3) and q^2=1(mod 3), hence
      > p^2+q^2-9=2(mod 3) is NEVER a square.

      That's what I meant to say.

      And p^2+2^2-9 = p^2-5 > (p-1)^2 for 2p-1>5 <=> p>3
      so this can't be a square, either.
      Which only leaves the trivial case q=3 => p^2 = p^2.

      The formulation is a quite obscure way to say something which has not much to do with primes.

      Maximilian
    • cino hilliard
      ... Generalizing to the n-th degree, with a touch of Serendipity, 1. p^n + q^n - b^n = k^2 has no integer solutions b, k for p, q prime, n odd 3. For p,q not
      Message 2 of 6 , Apr 25, 2009
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        >
        > When in doubt, Generalize.
        > p^2 + q^2 - b^2 = k^2 has infinite solutions for b prime, b*p>3 p prime and no solutions for b a multiple of 4 or 6, maybe finite solutions b = p*3 or p*2, p prime.
        > El Cino
        >

        Generalizing to the n-th degree, with a touch of Serendipity,

        1. p^n + q^n - b^n = k^2 has no integer solutions b, k for p, q prime, n odd > 3.

        For p,q not both prime, there are many solutions.



        p^n + q^n - b^n = k^m odd n > 3 has solutions but they are trivial, p, q = b.


        Maybe someone can prove 1.

        Cino







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