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  • Jon Perry
    I think I ve proved the 4in2 and 6in2 cases: http://www.users.globalnet.co.uk/~perry/maths/minierdosstrauss/minierdosstra uss.htm 5in2 seems more difficult.
    Message 1 of 1 , Jul 25, 2002
      I think I've proved the 4in2 and 6in2 cases:

      http://www.users.globalnet.co.uk/~perry/maths/minierdosstrauss/minierdosstra
      uss.htm

      5in2 seems more difficult.

      ---

      4/n in 2
      --------

      4/2k = 2/k

      4/(4k-1) - 4/4k = 4/(4k-1)4k

      If n is pure 1mod4, then:

      n(c+d)=4gcd

      implies LHS is 2mod4.

      6/n in 2
      --------

      6/2k = 3/k, which can be solved in 2 for all but k pure 1mod6

      6/3k = 2/k

      6/(3k-1) - 3/3k = (9k+3)/(3k-1)3k = 1/k + 2/k(3k-1)

      k odd implies 3k-1 is even.

      If n is pure 1mod6, then n(c+d)=6gcd fails.

      So 6/n has no solutions in 2 for pure 1mod6 and 2*(pure 1mod6)

      ---

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
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