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Re: [PrimeNumbers] OEIS A086381

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  • Phil Carmody
    ... Throw away the obfuscation of the squareroot - you want (p^2-q)/(p+1) to be an integer first, and and a square second. Trivially p^2-q = p^2+p-(p+q) So
    Message 1 of 2 , Sep 22, 2010
      --- On Thu, 9/23/10, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
      > Hello
      >
      > OEIS A086381 is:
      >
      > Numbers n such that n^2+2 and n^2+4 are primes.
      >
      > I have found that this secuence is the same:
      >
      > Numbers n such that n=((p^2-q)/(p+1))^(1/2) is integer.
      > p<q consecutive primes.

      Throw away the obfuscation of the squareroot - you want (p^2-q)/(p+1) to be an integer first, and and a square second.
      Trivially p^2-q = p^2+p-(p+q)
      So (p^2-q)/(p+1) = p - ((p+nextprime(p))/(p+1))
      We know nextprime(p) is near p, so, for p greater than not very much, that subtrahend is only ever going to be an integer when it's 2. I.e. q=p+2.

      And you've set p=n^2+2, so q=n^2+4.

      Q.E.D. (given that 'not very much' is bugger all)

      Too easy.

      Please try harder.

      Or don't.

      Phil
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