Comment from Benoit Cloitre (abcloitre(AT)wanadoo.fr), Aug 26 2002 A colleague supplied the following result: Let M be a symmetric n X n matrix with integer coefficients (in any commutative ring) satisfying m(i,j)=0 if i+j is even except when i=j=1 when m(1,1)=1 (1 can be replaced by any square), then the determinant of M is (in absolute value) always a perfect square. Sketch of proof: Consider the commutative ring A=Z[X_1,...,X_n] and let M_n be the matrix of a quadratic form on the rational fractions field of A. Let e_0,...e_n be the vectors for the canonical basis. Let b be the associated bilinear form. By hypothesis b(e_i,e_j)=0 if i+j is even except for b(e_0,e_0)=1 Let us consider the case n odd, n=2k+1 and let : f_0=e_n f_1=e_n+e_(n-2) f_2=e_n+e_(n-2)+e_(n-4) ........... f_k=e_n+e_(n-2)+....+e_1 f_(k+1)=e_(n-1) f_(k+2)=e_(n-1)+e_(n-3) ........... f_(n)=e_(n-1)+e_(n-3)+....+e_0 Then the matrix for b in the basis f_i is (by blocks) : N = 0------P t(P)-- Q where t(P) transposes P and Q is a matrix with all coefficients 0's except on the bottom right which is 1. Since P is a square matrix, the determinant of N is + or - a square (sign depending on n mod 4 ). But N is obtained from M via N=t(X)MX where X is the matrix from basis e to basis f. Hence, the determinant of M also is + or - a square in Q(X_1,...X_n). And this determinant is a polynomial of Z[X_1,...X_n], which is integrally closed, therefore the determinant of M is + or - a square in A=Z[X_1,...X_n]. For n even , we have the same form for N. P is not a square matrix. But the determinant of N is (except sign) the same than for the matrix obtained after deleting last row and last column of N. This way we are still in the case P square matrix.