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 <i>rationally congruent</i> if
there exists a matrix $S$ over the rationals such that $S^t A S = B$.

<b>Theorem</b> (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.