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.