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Proof of D. Andrica conjecture.

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  • John W. Nicholson
    http://www.primepuzzles.net/conjectures/conj_008.htm Proof of D. Andrica conjecture. Two squares one with area P(n) the other P(n+1) where P(n+1)
    Message 1 of 2 , Aug 2, 2002
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      http://www.primepuzzles.net/conjectures/conj_008.htm

      Proof of D. Andrica conjecture.

      Two squares one with area P(n) the other P(n+1) where P(n+1) < 2*P(n) by proof of "Bertrand's postulate" by Tschebycheff.

      Subtract the area of the square P(n)/P(n) = 1: So the area
      P(n+1)- P(n) = area D, the differance between the primes, is
      (P(n+1)-P(n))/P(n) < 2-1 < 1.

      So the sides have lengths of the squares are
      sqrt(P(n+1)) < sqrt(2*p(n))and sqrt(P(n)) < sqrt(P(n+1). Or after dividing by sqrt(P(n))
      1 < sqrt(P(n+1))/sqrt(P(n)) < sqrt(2)
      0 < sqrt(P(n+1))/sqrt(P(n))- 1 < sqrt(2)-1 < 1.

      The reason we can divide by sqrt(P(n)) is P(n+1) - P(n) = D is the differance between to squares.
      D = (sqrt(P(n+1))- sqrt(P(n))) (sqrt(P(n+1))+sqrt(P(n)))
      with
      2*sqrt(P(n)) < sqrt(P(n+1))+sqrt(P(n))
      making
      D /sqrt(P(n))~(sqrt(P(n+1))-sqrt(P(n)))*(2 *sqrt(P(n)))/sqrt(P(n))
      So QED

      Do y'all See any problems?



      [Non-text portions of this message have been removed]
    • richard042
      ... Maybe, Bertrand/Tschebycheff: pn
      Message 2 of 2 , Aug 3, 2002
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        --- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>
        >
        > Do y'all See any problems?

        Maybe,

        Bertrand/Tschebycheff: pn<(pn+1)<2*pn
        Take: sqrt(pn) < sqrt(pn+1) < sqrt(2*pn)
        divide by constant c: sqrt(pn)/c < sqrt(pn+1)/c < sqrt(2*pn)/c

        A.) subtract lhs: 0 < (sqrt(pn+1)-sqrt(pn))/c < (sqrt(2*pn)-sqrt
        (pn))/c

        Regardless of the value of c, A.) must hold.
        We need to get to Andrica's conjecture: sqrt(pn+1) - sqrt(pn)<1.
        So, multiply A.) by c to get:
        B.) 0 < sqrt(pn+1)-sqrt(pn) < sqrt(2*pn)-sqrt(pn)

        We can use B.) to prove Andrica's Conjecture only
        if it is always true that rhs <1 - i.e.:
        sqrt(2*pn)-sqrt(pn) < 1
        divide by c=sqrt(pn)to get:
        sqrt(2)-1 < 1/sqrt(pn)
        which is only true for n=1

        QESh..

        -Dick Boland
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