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For any integers $a,b,c\ge 0$, one has the well known identity or Dixon's Theorem: $$ \sum_{k\in\mathbb{Z}} (-1)^k \left(\begin{array}{c}a+b\\a+k\end{array}\right) \left(\begin{array}{c}b+c\\b+k\end{array}\right) \left(\begin{array}{c}c+a\\c+k\end{array}\right) \ =\ \frac{(a+b+c)!}{a!\ b!\ c!}\ . $$ There are tons of proofs in the literature, but I am wondering if there is one which only uses (possibly repeated) applications of the Chu-Vandermonde identity $$ \sum_{j\ge 0} \left(\begin{array}{c}a\\j\end{array}\right) \left(\begin{array}{c}b\\c-j\end{array}\right) \ =\ \left(\begin{array}{c}a+b\\c\end{array}\right)\ . $$

I should add, for more context, that I found such a proof and was wondering if something of that kind appeared in the literature. I would be very surprised if it didn't. A good example of what I mean by a proof only using Chu-Vandermonde is the derivation of the single index formula for 6j symbols by Racah in appendix B of his 1942 article "Theory of complex spectra II". BTW, appendix A of the same article also contains a proof of the above identity.


Edit (Mar 28, 2019):

The proof that I found is contained in Section 3 of the paper: "An algebraic independence result related to a conjecture of Dixmier on binary form invariants" which is now available. Given that I did not get an answer to this question since it was posted a few months ago, I would tend to think that my proof is new. Again, if you know otherwise, please point me towards the relevant reference.

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  • $\begingroup$ I don't know of such a proof in the literature. This question has now been open for some time, and I would be interested to see your proof using only Chu–Vandermonde. $\endgroup$ Aug 27, 2018 at 16:08
  • $\begingroup$ It's part of a proof or more general identities which I am including in a paper in preparation. I will definitely let you know when the paper is available. $\endgroup$ Aug 29, 2018 at 21:19
  • $\begingroup$ Is the paper in preparation now available? $\endgroup$
    – LSpice
    Feb 4, 2019 at 12:38
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    $\begingroup$ @MarkWildon: Sorry for the wait and thank you for your patience. The paper containing the said proof is now posted at arxiv.org/abs/1903.11147 $\endgroup$ Mar 28, 2019 at 13:51
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    $\begingroup$ @LSpice: Sorry for the wait and thank you for your patience. The paper containing the said proof is now posted. See my previous comment to Mark for the link. $\endgroup$ Mar 28, 2019 at 13:52

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