30 Jul
2016
30 Jul
'16
3:22 p.m.
Tom, So far it looks like, for prime m, n has a
rotationally symmetrical solution of order m if n is divisible by m^2.
And a rotationally symmetrical solution of order m^r if n is divisible by m^(r+1) (still assuming that m is prime), right? If we're looking at symmetrical solutions, we should probably include even values of n back in the game. For instance, when n=8 is there a solution with fourfold symmetry? Jim Propp