Source: Nikhil Garg (Sophomore, IITD) mailed them to me.
A checker starts at point (1,1). You can move checker using following moves :
1) if it is at (x,y) take it to (2x ,y ) or (x,2y)
2) if it is at (x,y) & x>y take it to (x-y,y)
3) if it is at (x,y) & x<y take it to (x,y-x)
Characterise all lattice points which can be reached.
You have a checker at (0,0) , (0,1) , (1,0), (2,0), (0,2), (1,1) each. You can make a move as follows:
if(x,y) is filled & (x+1,y) and (x,y+1) both are empty, remove checker from (x,y) & put one at each of (x+1,y) and (x,y+1)
Prove that under this move , you can not remove checker from all the six initial points.
Update (02/03/10): Solution posted by Nikhil Garg in comments!