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\begin{center}
\vskip 1cm{\LARGE\bf A Proof of Dixon's Identity \\
\vskip 1cm}
\large
Jovan Miki\'{c}\\
J.U. S\v{S}C ``Jovan Cviji\'{c}''\\
74480 Modri\v{c}a\\
Bosnia and Herzegovina\\
\href{mailto:jnmikic@gmail.com}{\tt jnmikic@gmail.com} \\
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\begin{abstract}
We give a proof of Dixon's binomial coefficient
identity using recurrence equations and
induction.
\end{abstract}

\section{Introduction}

In $1912$, Dixon \cite{AD} established the following famous identity 
\begin{equation}
\sum_{k=-a}^{a}(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}=\frac{(a+b+c)!}{a!\cdot b!\cdot c!}\mbox{,}\label{id:main} 
\end{equation}
where $a$, $b$, $c$ are nonnegative integers.

There are many proofs of Eq.~(\ref{id:main}).
In $1916$, MacMahon proved his master theorem. In 1962, Good \cite{IJG} found a short proof of Eq.~(\ref{id:main}) using MacMahon's master theorem.
Gessel and Stanton  \cite{DSIG} gave a short proof  using Laurent series in $ 1985$. 
In $1990$, Ekhad \cite{SE} gave a very short proof  using induction.
In $2003$, Guo \cite{VG} gave a short proof  using polynomials.

We give an elementary proof of Eq.~(\ref{id:main}) using recurrence
equations and induction. In order to obtain recurrence equations, we
will use some so-called auxiliary sums.

\section{Proof of Eq.~(\ref{id:main})}
\begin{proof}
Let $\mathbb{N} $ denote the set of positive integers, and let $\mathbb{N}_0 $ denote the set of nonnegative integers.
We let $S(a,b,c)$ denote the left side of Eq.~(\ref{id:main}), where
$a,b,c \in \mathbb{N}_0$.
We introduce the auxiliary sums $P(a,b,c)$, $Q(a,b,c)$ and $R(a,b,c)$ as follows
\begin{align}
P(a,b,c) &amp;= \sum_{k=-a}^{a}(-1)^k(a^2-k^2)\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}\mbox{,} \label{eq:2.1} \\
Q(a,b,c) &amp;= \sum_{k=-a}^{a}(-1)^k(b^2-k^2)\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}\mbox{,}\label{eq:2.2} \\
R(a,b,c)&amp; = \sum_{k=-a}^{a}(-1)^k(c^2-k^2)\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}\mbox{.}\label{eq:2.3}
\end{align}

Let $a,b,c \in \mathbb{N}$. We will use the well-known binomial identity $k\binom{n}{k}=n\binom{n-1}{k-1}$. Then 
\begin{align*}
P(a,b,c)&amp;=\sum_{k=-a+1}^{a-1}(-1)^k(a+k)\binom{a+b}{a+k}\binom{b+c}{b+k}(a-k)\binom{c+a}{a-k}\\
&amp;=\sum_{k=-a+1}^{a-1}(-1)^k(a+b)\binom{a+b-1}{a+k-1}\binom{b+c}{b+k}(c+a)\binom{c+a-1}{a-k-1}\\
&amp;=(a+b)(a+c)\sum_{k=-a+1}^{a-1}(-1)^k\binom{a-1+b}{a-1+k}\binom{b+c}{b+k}\binom{c+a-1}{c+k}\mbox{.}\\
\end{align*}
From the last equation above, it follows that
\begin{equation}
P(a,b,c)=(a+b)(a+c)S(a-1,b,c)\mbox{.}\label{eq:2.4} 
\end{equation}
 Similarly, we obtain that
\begin{equation}
Q(a,b,c)=(a+b)(b+c)S(a,b-1,c)\mbox{,}\label{eq:2.5} 
\end{equation}
and
\begin{equation}
R(a,b,c)=(a+c)(b+c)S(a,b,c-1)\mbox{.}\label{eq:2.6} 
\end{equation}
From Eqns.~(\ref{eq:2.1}) and (\ref{eq:2.2}), we have that 
\begin{equation}
P(a,b,c)-Q(a,b,c)=(a^2-b^2)S(a,b,c)\mbox{.}\label{eq:2.7}
\end{equation}
Let $a \ne b$. From Eqns.~(\ref{eq:2.4}), (\ref{eq:2.5}) and (\ref{eq:2.7}), it follows that
\begin{equation}
S(a,b,c)=\frac{1}{a-b}\bigl((a+c)S(a-1,b,c)-(b+c)S(a,b-1,c)\bigr)\mbox{.} \label{eq:2.8}
\end{equation}
Similarly, if $a \ne c  $, then it follows that
\begin{equation}
S(a,b,c)=\frac{1}{a-c}\bigl((a+b)S(a-1,b,c)-(b+c)S(a,b,c-1)\bigr)\mbox{.} \label{eq:2.9}
\end{equation}
We treat the case when $a=b=c$ separately.  
\begin{align*}
S(a,a,a)&amp;=\sum_{k=-a}^{a}(-1)^k{\binom{2a}{a+k}}^3\\
&amp;=\sum_{k=-a}^{a}(-1)^k\left(\binom{2a-1}{a+k}+\binom{2a-1}{a-1+k}\right)^3\\
&amp;=\sum_{k=-a}^{a}(-1)^k\left({\binom{2a-1}{a+k}}^3+3\binom{2a}{a+k}\binom{2a-1}{a+k}\binom{2a-1}{a-1+k}+{\binom{2a-1}{a-1+k}}^3\right)\\
&amp;=\sum_{k=-a}^{a}(-1)^k{\binom{2a-1}{a+k}}^3+\sum_{k=-a}^{a}(-1)^k{\binom{2a-1}{a-1+k}}^3+3S(a,a,a-1)\\
&amp;=\sum_{k=-a}^{a-1}(-1)^k{\binom{2a-1}{a+k}}^3+\sum_{k=-a+1}^{a}(-1)^k{\binom{2a-1}{a-1+k}}^3+3S(a,a,a-1)\\
&amp;=\sum_{k=-a}^{a-1}(-1)^k{\binom{2a-1}{a+k}}^3+\sum_{t=-a}^{a-1}(-1)^{t+1}{\binom{2a-1}{a+t}}^3+3S(a,a,a-1)\\
&amp;=3S(a,a,a-1)\mbox{.}
\end{align*}
Therefore,
 \begin{equation} S(a,a,a)=3S(a,a,a-1)\mbox{.}\label{eq:2.10} \end{equation}

Now we give a proof of Eq.~(\ref{id:main}) using induction:
 
Eq.~(\ref{id:main}) is true if $a=b=c=0$.
Let $n \in \mathbb{N}_{0} $ be fixed.
Assume that Eq.~(\ref{id:main}) holds for all nonnegative integers $a$, $b$ and $c$, such that $ a+b+c=n $.
 Let $a$, $b$ and $c$ be nonnegative integers such that $ a+b+c=n+1 $. Then we have three cases:

 \medskip
 
\noindent Case 1: If, at least, one of numbers $a$, $b$ or $c$ is equal to zero,
then Eq.~(\ref{id:main}) obviously holds. Therefore, we may assume that
$a,b,c \in \mathbb{N}$.

\medskip

\noindent Case 2: If $a=b=c$, then we use Eq.~(\ref{eq:2.10}). From the induction
hypothesis, we have that \[ S(a,a,a-1)=\frac{(3a-1)!}{a!\cdot a!\cdot
(a-1)!} \,\,\mbox{.}\]
 \begin{align*} \mbox{Then\,\,\,}
S(a,a,a)&amp;=3\cdot\frac{(3a-1)!}{a!\cdot a!\cdot (a-1)!}\\
 &amp;=3a\cdot\frac{(3a-1)!}{a!\cdot a!\cdot (a-1)!\cdot a}\\
  &amp;=\frac{(3a)!}{a!\cdot a!\cdot a!} \mbox {,\,\,as desired.}
 \end{align*}

\medskip

\noindent Case 3: If $ a\ne b$, then we use Eq.~(\ref{eq:2.8}). From the
induction hypothesis, we have that \[
S(a-1,b,c)=\frac{(a+b+c-1)!}{(a-1)!\cdot b!\cdot c!}\mbox{,
and}\,\,S(a,b-1,c)=\frac{(a+b+c-1)!}{a!\cdot (b-1)!\cdot c!}\mbox{.}\]
 \begin{align*}
 S(a,b,c)&amp;=\frac{1}{a-b}\bigl((a+c)S(a-1,b,c)-(b+c)S(a,b-1,c)\bigr)\\
 &amp;=\frac{1}{a-b}\bigl((a+c)\frac{(a+b+c-1)!}{(a-1)!\cdot b!\cdot
 c!}-(b+c)\frac{(a+b+c-1)!}{a!\cdot (b-1)!\cdot c!}\bigr)\\
 &amp;=\frac{1}{a-b}\cdot \frac{(a+b+c-1)!}{(a-1)!\cdot(b-1)!\cdot
 c!}\bigl(\frac{a+c}{b}-\frac{b+c}{a} \bigr)\\
  &amp;=\frac{(a+b+c-1)!\cdot (a^2-b^2+c(a-b))}{(a-b)\cdot a!\cdot b!\cdot
  c!}\\ &amp;=\frac{(a+b+c-1)!\cdot (a-b)(a+b+c)}{(a-b)\cdot a!\cdot
  b!\cdot c!}\\ &amp;=\frac{(a+b+c)!}{a!\cdot b!\cdot c!}  \mbox{,\,\,as
  desired.} \end{align*}

  If $a=b$ then $ a\ne c $. We do similarly as before using
  Eq.~(\ref{eq:2.9}).  This proves Case 3.  By induction,
  Eq.~(\ref{id:main}) follows.  
\end{proof} 

\section{Acknowledgments}
I want to thank to my professor Du\v{s}ko Joji\'{c}.

\begin{thebibliography}{9}

\bibitem{AD} A. C. Dixon, Summation of a certain series,
\textit{Proc.\ London Math.\ Soc.\ } \textbf{35} (1) (1903) 285--289.

\bibitem{SE} S. B. Ekhad, A very short proof of Dixon's theorem,
\textit{J.\ Combin.\ Theory}, Ser. A \textbf{54} (1990) 141--142.

\bibitem{DSIG} I. Gessel and D. Stanton, Short proofs of
Saalsch\"{u}tz's and Dixon's theorems, \textit{J.\ Combin.\ Theory},
Ser. A \textbf{38} (1985) 87--90.

\bibitem {IJG}  I. J. Good, Proofs of some `binomial' identities by
means of MacMahon's `Master Theorem' , \textit{Proc.\ Cambridge
Philos.\ Soc.\ } \textbf{58} (1962)  161--162.

\bibitem{VG} Victor J. W. Guo, A simple proof of Dixon's identity,
\textit{Discrete Math.} \textbf{268} (2003) 309--310.

\end{thebibliography}

\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary  11B65; Secondary 05A10.

\noindent \emph{Keywords:} recurrence equation, binomial coefficient,
Dixon's identity, auxiliary sum.

\bigskip
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\vspace*{+.1in}
\noindent
Received March 18 2016;
revised version received  May 10 2016.
Published in {\it Journal of Integer Sequences}, June 2 2016.

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\noindent
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