Following further trenchant observations from Dan, I conclude that introducing geometry at the start of this discussion may have been a mistake. [Not to mention committing a truly horrible typo in defining the matrices, over which I managed to confuse myself as much as everybody else --- I'm good at that!] So for now I'll just talk about polynomials with coefficients in some field --- usually \F_p for p = 2,3 in practice. My set of variables is X_(i,j), S_(i,j) where i,j in \Z, the subscript range being |i|+|j| <= n, with n remaining deliberately unspecified. My homogeneous quadratic forms are G_(i,j)(X) = X_(i,j) ^2 - X_(i-1,j) X_(i+1,j) - X_(i,j-1) X_(i,j+1), and the corresponding bilinear H_(i,j)(X,S) = 2 X_(i,j) S_(i,j) - X_(i-1,j) S_(i+1,j) - X_(i,j-1) S_(i,j+1) - S_(i-1,j) X_(i+1,j) - S_(i,j-1) X_(i,j+1). [We mustn't G_(i,j)(X) = (1/2)H_(i,j)(X,X) in case p = 2.] Now assign some appropriate numerical values to S, and consider the set J(S) = {X | G_(i,j)(X) = H_(i,j)(X,S) = G_(i,j)(S) = 0} for all |i|+|j| < n. I can prove that the dimension of J(S) equals 2. [Dimension is defined here algebraically, as n-1 less the least cardinal of any basis of the ideal whose zero set defines the algebraic variety.] Taking p = 3 for instance, examination of actual values assigned to S suggests that the number of "points" in J(S) is usually 24. [Points are solutions, not all zero, and modulo constant factors --- {-1,+1} in the case of p = 3.] Question: is it possible to actually prove this sort of result; or otherwise to bound the number of points in J(S)? Third time lucky, I hope! Fred Lunnon