A sum of sum involving binomial coefficent.(my conjecture)

Question

While using MATLAB I found out that $\displaystyle\sum_{k=0}^{n}\frac{(\sum_{r=\max (0,k+m-n)}^{\min (k,m)}(-1)^{r}\binom{n-m}{k-r}\binom{m}{r})^{2}}{\binom{n}{k}}=\frac{2^{n}}{\binom{n}{m}} $ hold for all $m\leq n\leq 100\ $ and I proved that the equality hold for all $m=0,1,n-1,n\ $ but I couldn't prove it in general (or disprove).
Any help is appreciated! Thanks in advance.

Answer 1

The credit for this proof goes to Dan Carmon.

Let $V$ be the space of homogeneous polynomial of degree $n$ in 2 variable $x$ and $y$ and let $T$ be the linear map defined as: $T(p)(x,y)=p(x+y,x-y)$.

Now note that: $T^{2}(p)(x,y)=T(p)(x+y,x-y)=p((x+y)+(x-y)),(x+y)-(x-y))=p(2x,2y)=2^{n}p(x,y)$

Let $A$ be the matrix representation of the linear transformation $T$ with the base $\left \{ x^{n},x^{n-1}y,...,xy^{n-1},y^{n} \right.\left. \right \}$

$\sum_{k=0}^{n}a_{m,k}x^{n-k}y^{k}=T(x^{n-m}y^{m})=(x-y)^{m}(x+y)^{n-m}\\=\sum_{k=0}^{n}\sum_{r=max(0,k+m-n)}^{min(m,k)}(-1)^{r}\binom{m}{r}\binom{n-m}{k-r}x^{n-k}y^{k}$

We are getting closer to end because now we know that $a_{m,k}=\sum_{r=max(0,k+m-n)}^{min(m,k)}(-1)^{r}\binom{m}{r}\binom{n-m}{k-r}$

and with sum (some) manipulation $\binom{m}{r}\binom{n-m}{k-r}\binom{n}{m}=\frac{m!(n-m)!n!}{r!(m-r)!(k-r)!(n-m-k+r)!(m-n)!m!}\\=\frac{n!}{r!(m-r)!(k-r)!(n-m-k+r)!}=\frac{n!(n-k)!k!}{r!(m-r)!(k-r)!(n-m-k+r)!k!(n-k)!}=\binom{k}{r}\binom{n-k}{m-r}\binom{n}{k}$

So $\binom{n}{m}a_{m,k}=\binom{n}{k}a_{k,m}$

The identity we seek to prove is equivalent to $\sum_{k=0}^{n}\frac{\binom{n}{m}a_{m,k}^{2}}{\binom{n}{k}}=2^{n}$

but the last identity yield:

$\sum_{k=0}^{n}\frac{\binom{n}{m}a_{m,k}^{2}}{\binom{n}{k}}=\sum_{k=0}^{n}a_{m,k}a_{k,m}=\left [ A^{2} \right ]_{m,m}=\left [ 2^{n} I\right ] _{m,m}=2^{n} $

$\blacksquare$