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 $Q_i$, 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$Q\left(t\right)=t^{n-1}$(I suspect the below proof works with any polynomial$Q$of degree$n-1$(not less!), but I'm not completely sure and don't have the time to check) and let$Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$for every$i\in\mathbb Z$. Being a polynomial in$i$of degree$2\left(2\left(n-1\right)+1\right)=4n$2\left(2\left(n-1\right)+1\right)<4n$ (for fixed $t$ and $s$), the term $Q_i\left(t\right)Q_i\left(s\right)$ satisfies $$\sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0,$$ since the $\left(4n+1\right)$-th 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$n$pairwise 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$t=0$, and get a contradiction because Vandermonde's determinant is nonzero. 3 added 61 characters in body; added 300 characters in body 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$Q_i$, so that your equation $$\sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)^2 sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)^2 P_i(t)\hat P_i(s)$$ becomes $$\sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0. sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0.$$ Now let$Q$be the polynomial$Q\left(t\right)=t^{n-1}$(I suspect the below proof works with any polynomial$Q$of degree$n-1$(not less!), but I'm not completely sure and don't have the time to check) and let$Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$for every$i\in\mathbb Z$. Being a polynomial in$i$of degree$2\left(2\left(n-1\right)+1\right)=4n$(for fixed$t$), t$ and $s$), the square term $Q_i\left(t\right)^2$ Q_i\left(t\right)Q_i\left(s\right)$satisfies $$\sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0, sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0,$$ since the$\left(4n+1\right)$-th finite difference of a polynomial of degree$4n$is zero. Due to the symmetry of the function$i\mapsto Q_i\left(t\right)^2$Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$\sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0. sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0.$$ 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 $n$ pairwise 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 $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.
Wadim, isn't that 95% of the proof? Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_i$, so that your equation $$\sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)^2 =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)^2$$ becomes $$\sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0.$$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$ (I suspect the below proof works with any polynomial $Q$ of degree $n-1$ (not less!), but I'm not completely sure and don't have the time to check) and let $Q_i\left(t\right)=\left(2n-i\right)^2Q\left(t-\left(2n-i\right)^2\right)$ Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$for every$i\in\mathbb Z$. Being a polynomial in$i$of degree$2\left(2\left(n-1\right)+1\right)=4n$(for fixed$t$), the square$Q_i\left(t\right)^2$satisfies $$\sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0,$$ since the$\left(4n+1\right)$-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(t\right)^2$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$\sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0.$$ 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 $n$ pairwise 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 $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.