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• 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.

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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)

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Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
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