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Re: [PrimeNumbers] 4n^2-3n-1 and 4n^2+3n-1

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  • Jack Brennen
    4n^2-3n-1 == (n-1)(4n+1) 4n^2+3n-1 == (n+1)(4n-1) Only way either one is prime is if one of the two cofactors is 1 or -1 and the other one is a prime. You can
    Message 1 of 2 , Mar 24, 2010
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      4n^2-3n-1 == (n-1)(4n+1)
      4n^2+3n-1 == (n+1)(4n-1)

      Only way either one is prime is if one of the two cofactors is 1 or -1
      and the other one is a prime. You can quickly check all of those
      possibilities, and eliminate them all; thus, no primes of either form.



      Sren wrote:
      > Those two formulas I found while I was working with exclusion lines in Ulams prime spiral. None of them have primes at least up to n=10000.
      >
      > Are there any who have an explanation ?
      >
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