The goal of the problem is to determine (or merely bound) F(n), which is the maximum possible number of n-vectors over Z3, such that no three sum to 0. More precisely such that for any three, call them x,y,z, it can only happen that x+y+z=0 if x=y=z. As my first trivial remark... obviously, if it is the case that no three of the vectors x,y,z exhibit ANY linear relation a*x+b*y+c*z=0 over Z3, (a,b,c)!=(0,0,0), that suffices. For that, it in turn suffices if our set of (column) n-vectors, n>3, happens to form an "orthogonal array of strength 3." For that, it in turn suffices if the set of vectors is the dual of a hamming-distance-4 linear code over GF3. So it seems to me that by using the "duals of the extended ternary Hamming codes" we obtain a proof that F( (3^r-1)/2 + 1 ) >= 3^r. Oh. Well, that was a quite weak result, F(n) >= 2*n-1 for an infinite set of n. Supposedly it is known that F(n) > 2.21^n, a far stronger lower bound. Still, since Hamming codes are "perfect" my bound is presumably best possible using its technique. So this demonstrates that demanding all 26 linear relations be impossible causes a tremendously different effect than only demanding (what the problem calls for) it for the 7 relations (a,b,c) = (1,1,1) and perms(0,1,2). This is a very interesting problem. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)