- I think I've proved the 4in2 and 6in2 cases:
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:
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)