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Andrica's conjecture

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  • Matteo Vitturi
    Good evening. At http://www.primepuzzles.net/conjectures/conj_008.htm I ve read Jim Fougeron s comment about Andrica s conjecture. I m not a mathematician, so
    Message 1 of 2 , Jun 4, 2013
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      Good evening.

      At http://www.primepuzzles.net/conjectures/conj_008.htm I've read Jim Fougeron's comment about Andrica's conjecture. I'm not a mathematician, so please be benevolent with me, but his proof attempt led me the following reasoning:

      Write Andrica's Conjecture

      sqrt( p[n+1] ) - sqrt( p[n] ) < 1

      in the gap form

      g[n] < 2sqrt( p[n] ) + 1

      Choose y such that q^2 is the square number the nearest to and less than p[n]

      q^2 + y := p[n]
      g[n] < 2sqrt( q^2 + y ) + 1

      Now, there must exist a number w such that

      2sqrt( q^2 + y ) + 1 = 2q + 2w + 1

      Let us ignore 2w for now, and write a stronger condition

      g[n] < 2q + 1

      We note that 2q + 1 is the difference between (q+1)^2 and q^2.
      If this inequality is true, then starting counting at square q^2 and fetching odd integers until a prime is encountered it is very likely to find a prime before the next square (q+1)^2.

      Using logarithmic integral fuction to find how many prime should we find between two square numbers

      N(q) := Li((q+1)^2) - Li(q^2)

      it can be seen that N(q) is monotonically increasing and at least two primes are expected between any two squares that asymptotically makes true the stronger condition and, a fortiori, Andrica's conjecture.

      Does that resemble a sound proof?

      Regards.

      _ Matteo Vitturi.



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    • WarrenS
      ... --it is a reasonable heuristic argument, nowhere near a proof.
      Message 2 of 2 , Jun 4, 2013
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        > Does that resemble a sound proof?

        --it is a reasonable heuristic argument, nowhere near a proof.
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