On an old post related to determinants

[benoit]

 From  "Algebragirl" (a Jussieu member), here an answer to a question 
posted by Santi Spadoro few month ago, concerning :

  http://www.research.att.com/projects/OEIS?Anum=A069191

Seems to me that the following  answer is appropriate since there is in 
fact a more general result. This problem is not directly related to 
prime numbers.  I remember Michael Somos gave an answer which was 
probably equivalent but expressed with pari-gp codes.

Let M be a nxn matrix with integer coefficients (can be in any 
commutative ring) satisfying m(i,j)=0 if i+j is even  except when i=j=1 
  where 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 (sorry for the unelegant translation) :

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. From hypothesis : b(e_i,e_j)=0 if i+j is even 
except 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 mod4 ). 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 "intégralement clos", 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.


Benoit Cloitre
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