I have stumbled upon the following statement and have verified it computationally for many $n$ (up to n=500, it took a long time for my computer to do out all of the math), yet I have no idea how to go about proving it.
$$\sum _{ i=0 }^{ n } \sum _{ j=0 }^{ n } \frac { { ((2n)! })^{ 2 }(-1)^{ i+j }(2i+2j)! }{ (n-i)!(n-j)!(2i)!(2j)!2^{ i+j }(i+j)! } =\frac { (4n)! }{ { 4 }^{ n }(2n)! } $$
One thing to note is that if we multiply the top and bottom of the fraction by $(i)!$, $(j)!$, and $(n!)^2$, then certain terms simplify into the factorial definition of the binomial coefficient, so the sum is also expressible as:
$$\sum _{ i=0 }^{ n } \sum _{ j=0 }^{ n } (-1)^{ i+j } \binom{n}{i}\binom{n}{j} \frac { { ((2n)! })^{ 2 }}{(n!)^2}\frac{(i)!}{(2i)!}\frac{(j)!}{(2j)!}\frac{(2i+2j)! }{2^{ i+j }(i+j)! } =\frac { (4n)! }{ { 4 }^{ n }(2n)! } $$
Also, let us define $P(n)$ to be equal to the product of the first $n$ positive odd numbers (or more specifically, $P(n)=(2n-1)!!$). We can use the explicit formula $P(n)=\frac{(2n)!}{2^n n!}$ (see here for explanation) to rewrite various expressions:
$$\frac{(2i+2j)! }{2^{ i+j }(i+j)! }=P(i+j)$$
$$\frac{((2n)!)^2}{(n!)^2}=(P(n))^2 2^{2n}$$
$$\frac{(i)!}{(2i)!}=\frac{1}{2^i P(i)}$$
$$\frac{(4n)!}{4^{n}(2n)!}=P(2n)$$
And so the statement can be rewritten as:
$$ \sum _{ i=0 }^{ n } \sum _{ j=0 }^{ n } (-1)^{ i+j } \binom{n}{i}\binom{n}{j} \frac{2^{2n}(P(n))^2 P(i+j)}{2^{i+j}P(i)P(j)}=P(2n) $$
Interestingly enough, this proves that the left hand side must be an integer (see here for an explanation), which i believe is a step in the right direction.
This is all just random stuff I have been trying to attempt to prove the identity although so far I have not gotten very far. If anyone has any idea on how I could prove this identity or even just begin to prove the identity, please help me out! Any and all help is appreciated!
Edit, here is some more work done on the topic:
Let us define: $$R_n = \sum _{ i=0 }^{ n } \sum _{ j=0 }^{ n } \frac { { ((2n)! })^{ 2 }(-1)^{ i+j }(2i+2j)! }{ (n-i)!(n-j)!(2i)!(2j)!2^{ i+j }(i+j)! }$$
I am currently trying to prove that $R_n=P(2n)$. Since $P(n)$ is the product of the first $n$ odd integers, and $2n-1$ is the $n$th odd integer, it follows that $P(2(n+1))=P(2n+2)=P(2n) \cdot (4n+1) \cdot (4n+3)$. Thus (since the base case works, and you can confirm that if you truly want), a statement equivalent to what I am trying to prove is:
$$R_{n+1}=(4n+1)(4n+3)R_{n}$$
Best Answer
Consider the polynomial $$ f_n(x)=\left(x^2+1-\frac{\left(x+1\right)^2}{2}\right)^{2n}. $$ The left and right hand sides of the OP's identity each compute the coefficient of $x^{2n}$ in $f_n(x)$.
One the one hand, we can expand $f_n$ using the trinomial and binomial theorems: \begin{align*} f_n(x)&=\sum_{a+b\leq 2n}\frac{(2n)!}{a!b!(2n-a-b)!}x^{2a}\left(x+1\right)^{2(2n-a-b)}\frac{(-1)^{2n-a-b}}{2^{2n-a-b}}\\ &=\sum_{\substack{a+b\leq 2n\\r\leq 4n-2a-2b}}\frac{(2n)!}{a!b!(2n-a-b)!}{4n-2a-2b\choose r}\frac{(-1)^{2n-a-b}}{2^{2n-a-b}}x^{2a+r}. \end{align*} To get $x^{2n}$, we need $r=2n-2a$, so that $a\leq n$. Additionally, $r=2n-2a$ and $r\leq 4n-2b-2b$ imply $b\leq n$, so we are summing over $0\leq a,b\leq n$. Make the substitution $i=n-a$, $j=n-b$ (so that $r=2i$). In terms of $i$ and $j$, the sum is over $0\leq i,j\leq n$. This shows that the coefficient of $x^{2n}$ is the left hand side of the OP's identity.
On the other hand, we can simplify $f_n$: \begin{align*} f_n(x)&=\left(\frac{(x-1)^2}{2}\right)^{2n}\\ &=\frac{(x-1)^{4n}}{4^n}. \end{align*} The binomial theorem then shows the coefficient of $x^{2n}$ is the right hand side of the OP's identity.