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Re: abc|f(a)f(b)f(c) (was n^4+4)

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  • Mike Oakes
    ... If a,b,c are not required to be primes, then there are many solutions. If a f(a)=931=7^2*19 b=133=7*19 =
    Message 1 of 11 , Jan 31, 2007
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      --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
      wrote:
      >
      > Let f(x)=x^2-x+1.
      > Find 2 triplets, {a,b,c}, of primes
      > such that abc|f(a)f(b)f(c).
      >

      If a,b,c are not required to be primes, then there are many solutions.

      If a<b<c, the one with minimum a is:
      a=31 => f(a)=931=7^2*19
      b=133=7*19 => f(b)=17557=97*181
      c=181 => f(c)=32581=31*1051

      -Mike Oakes
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