Rational congruence of binomial coefficient matrices

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ is, where $t$ and $s$ are two independent variables.

Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_{2i+1}$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$. (With some work, the proof below works just as well if $Q$ is any polynomial of degree $n-1$ (not less!), but let me use $t^{n-1}$ for simplicity's sake.) Let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. For any fixed $t$ and $s$, the term $Q_i\left(t\right)Q_i\left(s\right)$ is a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)<4n$, and thus satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0, $$ since the $4n$-th finite difference of a polynomial of degree $< 4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$ (or alternatively, just take their coefficients), and get a contradiction because Vandermonde's determinant is nonzero.

Expecting to be criticised, I nevertheless try to explain an approach to the problem (for the first case). Assume that there exist two family of polynomials $P_i(t)$ and $\hat P_i(t)$, $i=0,1,\dots,n-1$, each spanning the space of polynomials of degree less than $n$ such that $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)^2 =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)^2 $$ identically in $t$; this, for example, happens if the both sides are equal for $n$ distinct points. Then replacing the monomials $t^0,t^1,\dots,t^{n-1}$ by variables $t_0,t_1,\dots,t_{n-1}$ we obtain the linear forms $L_i(t_0,t_1,\dots,t_{n-1})$ and $\hat L_i(t_0,t_1,\dots,t_{n-1})$ such that $L_i(1,t,\dots,t^{n-1})=P_i(t)$ and $\hat L_i(1,t,\dots,t^{n-1})=\hat P_i(t)$, $i=0,1,\dots,n-1$, and $$ \sum_{i=0}^{n-1}\binom{4n}{2i}L_i(t_0,t_1,\dots,t_{n-1})^2 =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat L_i(t_0,t_1,\dots,t_{n-1})^2, $$ the desired equivalence. So far, I could only manage some examples of the polynomial expansions, like $$ \frac{(1+t)^{4n}+(1-t)^{4n}}{2t^{2n}} =\sum_{i=0}^{2n}\binom{4n}{2i}t^{2i-2n} =\sum_{i=0}^{n-1}\binom{4n}{2i}(t^{-2(n-i)}+t^{2(n-i)})+\binom{4n}{2n} $$ $$ =\sum_{i=0}^{n-1}\binom{4n}{2i}(t^{n-i}-t^{-(n-i)})^2+\sum_{i=0}^{2n}\binom{4n}{2i} $$ $$ =\biggl(t-\frac1t\biggr)\sum_{i=0}^{n-1}\binom{4n}{2i}P_i\biggl(t+\frac1t\biggr)^2+2^{4n-1}, $$ in other words, $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i\biggl(t+\frac1t\biggr)^2 =\frac{(1+t)^{4n}+(1-t)^{4n}-2^{4n}t^{2n}}{2t^{2n}(t-1/t)}. $$ So, the question is whether we can write the rational function on the right-hand side as $$ \sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i\biggl(t+\frac1t\biggr)^2. $$

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.