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24290Re: [PrimeNumbers] Another form of probabilistic test for primeness

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  • Kermit Rose
    Jun 24, 2012
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      On 6/23/2012 5:44 PM, Peter Kosinar wrote:
      >> It will be a matter of empirical research to determine lower bounds
      >> on composite y such that
      >>
      >> 2^(3y+1) = 2^(y+3) mod (3y)
      >> 2^(5y+1) = 2^(y+5) mod (5y)
      >> 2^(7y+1) = 2^(y+7) mod (7y)
      >
      > Four shalt be the number thou shalt count, and the number of the
      counting shalt be four.
      >
      > For any odd prime x, we have 2^(4x+1) = 2^(5 + 4(x-1)), so 2^(4x+1) =
      0 (mod 4) and 2^(4x+1) = 2^5 (mod x), thus by Chinese Remainder Theorem,
      we get 2^(4x+1) = 2^5 (mod 4x)
      >
      > We also have 2^(x+4) = 2^(5 + (x-1)), which gets us
      > 2^(x+4) = 0 (mod 4) and 2^(x+4) = 2^5 (mod x). Combining them by CNT
      > gives 2^(x+4) = 2^5 (mod 4x)
      >
      > Ergo, 2^(4x+1) = 2^(4+x) mod (4x), and 4 is the smallest
      counterexample for any odd prime x.
      >
      > Peter

      Hello Peter.

      Thanks.

      My test is even worse than you have discovered.

      For any composite y such that 2^y = 2 mod y, (Original Fermat test
      which fails too many times)
      and for any odd prime x,

      Calculate 2^(xy-x-y+1) mod x and 2^(xy-x-y+1) mod y

      2^(xy-x-y+1) = (2^x)^y * 2^(-x-y+1) = 2^y * 2^(-x-y+1) = 2^(y-x-y+1) =
      2^(-x+1) = 2^(1-1) mod x


      2^(xy-x-y+1) = (2^x)^y * 2^(-x-y+1) = 2^x * 2^(-x-y+1) = 2^(x-x-y+1) =
      2^(-y+1) = 2^(1-1) mod y


      Thus by chinese remainder theorem, 2(xy-x-y+1) = 1 mod (xy).

      This proves that my proposed "new" test is only a disguised Fermat test.

      Kermit Rose
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