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\centerline
{\bf{A WZ PROOF OF A ``CURIOUS'' IDENTITY}}

\rm
\bigskip
\centerline{ {\it 
Shalosh B. Ekhad   and  Mohamud Mohammed
} \footnote{$^1$}
{\eightrm  \raggedright
Department of Mathematics, Rutgers University,
Piscataway, NJ 08854, USA. 
%\break
{\eighttt mohamudm@math.rutgers.edu} \hfill \break
{\eighttt http://www.math.rutgers.edu/\Tilde mohamudm/.}
April 12, 2002. 
{\eighttt}
} 
}

{\bf Abstract:} We give a short WZ-proof of an identity that appeared recently in Integers.\bigbreak
In [1] and [2], the identity, 
$$ \sum_{i=0}^{m}(x+m+1) {(-1)^i {{x+y+i} \choose {m-i}} {{y+2i} \choose {i}}}
 -\sum_{i=0}^{m} {{{x+i} \choose {m-i}}(-4)^i}=(x-m) {{x} \choose {m}}...........(*)$$
was proved using generating functions and double recursions respectively.

Here we cleverly construct the function 
$$ g(i) = {-{{{x} \choose {i}}(x^2-2ix-x+i+i^2)} \over
 {2(1+i)(x+i+2)(x+i+1)(-2)^i}}$$ with the motives that

$$\sum_{i=0}^{m} {{{x+i}\choose {m-i}}} (-4)^i = (-2)^{m}(x+m+1) \left ( {{1} \over {x+1}}+ \sum_{i=0}^{m-1}g(i) \right )
..........(**) $$

The first sum, call it $T(m)$,  on the LHS of $(*)$ divided by $(x+m+1)$ satisfies the following recurrence equation as found by EKHAD.

$   a_{0}(m)T(m)+a_{1}(m)T(m+1)+a_{2}(m)T(m+2)+a_{3}(m)T(m+3)=0,$ where

$ a_{0}(m)= 2(x-m-1)(x-m-2) ; a_{1}(m)=-\left (x-m-2\right )\left (2\,y-x+5\,m+11 \right) $,

$ a_{2}(m)= \left (-yx+3\,ym-2\,xm +4\,{m}^{2}+8\,y-5\,x +21\,m+28\right); a_{3}(m)=\left (m+3\right )\left (y+m+3\right) $

Now, it is routine(use maple!) to check that the sum of the RHS's of $(*)$ and $(**)$ divided by $(x+m+1)$ satisfies the recurrence equation. Moreover, 
both sides agree for $ m=0,1,2 $.   Q.E.D.
 

{\bf REFERENCES}
   
[1] Alois Panholzer, and Helmut Prodinger. A generating functions proof of a curious identity. Integers, pages A6, 3 pp. (electronic), 2002.

[2] Zhi-Wei Sun. A curious identity involving binomial coefficients. Integers, pages A4, 8 pp. (electronic), 2002

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