Timeline for Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$

10 events

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Oct 17, 2017 at 8:15	comment	added	Fedor Petrov		of course it is true: $(f/f_0)'=((f\cdot h)/(f_0\cdot h))'=((f\cdot h)'f_0-(f_0\cdot h)'f)/f_0^2h^2=0$, where $h=1-2(x+y)+(x-y)^2$, thus $f/f_0=\text{const}=1$.
Oct 17, 2017 at 5:38	comment	added	ken		Is it true that $f(0)＝f_0(0),(f(1-2(x+y)+(x-y)^2))_x'＝(x-y-1)f,(f_0(1-2(x+y)+(x-y)^2))'＝(x-y-1)f_0⇒f＝f_0？$
Oct 16, 2017 at 19:10	comment	added	Max Alekseyev		The identity $f(x,y)=f_0(x,y)$ can be also obtained by Lagrange Inversion. Or, derived from the g.f. $F(x,y)$ given in my answer as $f(x,y)=F(x,\frac{y}{x})$.
Oct 16, 2017 at 15:33	comment	added	Fedor Petrov		@Vincent look at a coefficient of $x^{2a+1}y^{2b+1}$ in $xyf(x^2,y^2)f(x^2,y^2)$. It is nothing but the left hand side of the desired identity.
Oct 16, 2017 at 15:08	comment	added	Vincent		Can you please explain how $xyf^2(x^2, y^2)$ being an odd part of $h(x, y)$ relates to the question being asked?
Oct 16, 2017 at 12:47	history	edited	Fedor Petrov	CC BY-SA 3.0	deleted 46 characters in body
Oct 16, 2017 at 11:02	comment	added	Shahrooz		Nice one dear Petrov!
Oct 16, 2017 at 10:40	history	edited	Fedor Petrov	CC BY-SA 3.0	added 18 characters in body
Oct 16, 2017 at 10:13	history	edited	Fedor Petrov	CC BY-SA 3.0	added 220 characters in body
Oct 16, 2017 at 7:00	history	answered	Fedor Petrov	CC BY-SA 3.0