TITLE: Rational congruence of binomial coefficient matrices
QUESTION [13 upvotes]: Skip Garibaldi asks if there is an elementary proof of the following fact that "accidentally" fell out of some high-powered machinery he was working on.
Say that two matrices $A$ and $B$ over the rationals are rationally congruent if
there exists a nonsingular matrix $S$ over the rationals such that $S^t A S = B$.
Theorem (Garibaldi).  Suppose $n \equiv 0 \pmod 4$. Then the diagonal matrices
$$A = diag\left[\binom{n}{0}, \binom{n}{2}, \binom{n}{4}, \ldots, \binom{n}{n/2 - 2}\right]$$
and
$$B = diag\left[\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots, \binom{n}{n/2 - 1}\right]$$
are rationally congruent. Similarly, suppose $ n \equiv 2 \pmod 4$. Then the matrices
$$A = diag\left[\binom{n}{0}, \binom{n}{2}, \binom{n}{4}, \ldots, \binom{n}{n/2 - 1}\right]$$
and
$$B = diag\left[\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots, \binom{n}{n/2 - 2}, \frac{1}{2}\binom{n}{n/2}\right]$$
are rationally congruent.

REPLY [3 votes]: There is a problem with this approach (I refer to Wadim Zudilin's answer). At least I don't see how to get from the first displayed equation (involving the $P_i$) to the second (involving the $L_i$) in his post.
Here is an example:
$(t^2+t-2)^2 + (3t+2)^2 + 1^2 = (t^2+t+2)^2 + (t+2)^2 + 1^2$
but 
$(t_2+t_1-2t_0)^2 + (3t_1+2t_0)^2 + t_0^2 \ne (t_2+t_1+2t_0)^2 + (t_1+2t_0)^2 + t_0^2$,
by considering the coefficient of $t_0t_2$ or of $t_1^2$.
Clearly, if the second equation were true it would imply the first. But the problem with the converse is that the first involves 5 degrees of freedom (the coefficients of $t^i$ with $0 \le i \le 4$), whereas the second has 6 (the coefficients of the $t_it_j$ with $i \le j$).
Edit: Here is a better example:
$(t^2+1)^2 -2 (t)^2 + 1^2 = (t^2-1)^2 + 2 (t)^2 + 1^2$,
but the two forms $diag(1,-2,1)$ and $diag(1,2,1)$ don't even have the same signature.