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RE: [PrimeNumbers] Quiz day soon

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  • Paul Leyland
    Here is a solution in integers for which b,c,n, x, y, and z are all different. There are an infinite number of solutions, of course. b = -1 c = 1 n = 0 x =
    Message 1 of 3 , Dec 4, 2003
      Here is a solution in integers for which b,c,n, x, y, and z are all different. There are an infinite number of solutions, of course.

      b = -1
      c = 1
      n = 0
      x = -16
      y = 2
      z = 81


      Paul


      > -----Original Message-----
      > From: paulmillscv [mailto:paulmillscv@...]
      > Sent: 04 December 2003 14:25
      > To: primenumbers@yahoogroups.com
      > Subject: [PrimeNumbers] Quiz day soon
      >
      > Hello to Primers,
      > Xmas Quiz day is Monday December 15th. The quiz link will be
      > posted then and available on all time zones on the 16th. The
      > questions will be difficult but not without huge clues as to the
      > solution. As a warm up can you find a solution (non-trivial) to this
      > question.
      >
      > Find 3 numbers in Arithmetic Progression A=cn+b B=c(n+1) +b and C=c
      > (n+2) +b so that xA^3 , yB^3 and zC^3 are all FOURTH powers. I.e.
      > xA^3 = D^4 yB^3 = E^4 and zC^3 = F^4. All letters have integer
      > values.
      >
      > That should keep you occupied until 15th December!
      >
      > Regards,
      >
      > Paul Mills
      > Kenilworth,
      >
      >
      >
      >
      >
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    • elevensmooth
      I m still trying to figure out what consitutes a non-trivial solution. x=A y=B z=C is a solution for any choices of A, B, C - having nothing to do with their
      Message 2 of 3 , Dec 5, 2003
        I'm still trying to figure out what consitutes a non-trivial solution.

        x=A
        y=B
        z=C

        is a solution for any choices of A, B, C - having nothing to do with
        their arithmetic progression. That's surely trivial.

        Perhaps the x, y, and z are supposed to be given as fixed values?
        That would the problem more difficult.

        --- In primenumbers@yahoogroups.com, "paulmillscv" <paulmillscv@y...>
        wrote:
        > Hello to Primers,
        > Xmas Quiz day is Monday December 15th. The quiz link will be
        > posted then and available on all time zones on the 16th. The
        > questions will be difficult but not without huge clues as to the
        > solution. As a warm up can you find a solution (non-trivial) to this
        > question.
        >
        > Find 3 numbers in Arithmetic Progression A=cn+b B=c(n+1) +b and C=c
        > (n+2) +b so that xA^3 , yB^3 and zC^3 are all FOURTH powers. I.e.
        > xA^3 = D^4 yB^3 = E^4 and zC^3 = F^4. All letters have integer
        > values.
        >
        > That should keep you occupied until 15th December!
        >
        > Regards,
        >
        > Paul Mills
        > Kenilworth,
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