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\bf
\centerline{How Borwein and Bradley's Hypergeometric Ugly Duckling
Turned Into a Beautiful}
\centerline
{`Eulerian' Swan, thanks to Almkvist and Granville's
Magic}
\it
\medskip
\centerline{Doron ZEILBERGER\footnote{$^1$}
{\baselineskip=9pt
\eightrm  \raggedright
Department of Mathematics, Temple University,
Philadelphia, PA 19122, USA. 
{\baselineskip=9pt
\eighttt zeilberg@math.temple.edu ; \break http://www.math.temple.edu/\Tilde 
zeilberg. }
Supported in part by the NSF. Dec. 12, 1996. Exclusive to
the author's WWW site.
}}
\medskip
\qquad
{\it Every Mathematician has only a few\footnote{$^2$}
{\baselineskip=9pt
\eightrm  \raggedright
However, `few' should be at least 2.
}
 tricks}---
{\rm Gian-Carlo Rota [Notices 44(1)(1/1997), 24].}
\medskip
\rm
\bigskip
Last spring, Jon Borwein and David Bradley[BB] came up with
an intriguing conjecture. They stated several equivalent statements,
one of them, due to Wenchang Chu, states:
$$
\sum_{k=1}^{n}{{2n^2} \over {k^2}}
{{\prod_{i=1}^{n-1}{(i^4+4k^4)}} \over
{\prod_{{{i=1} \atop {i \neq k}}}^{n}{(k^4-i^4)}} }=
{{2n} \choose {n}} \quad .
\eqno(BBC)
$$
 
Superficially, this looks like grit for the EKHAD mill.
A closer look, however, shows that it is not
doable {\it as is}, because  $(k^4-i^4)$ and
$(i^4+4k^4)$ don't factor completely over the rational numbers.
Nevertheless, 
since I had this nifty hammer, I was sure that this identity
is  yet another nail, that just needs some polishing before it can be hit.
I tried lots of Rube Goldberg contraptions, but
to no avail, EKHAD refused to find the `trivializing generalization'.
 
My mistake was in being trapped in the hypergeometric paradigm.
Gert Almkvist and Andrew Granville's 
beautiful proof[AG] showed
that the hypergeometric appearance is but a red herring. Instead it followed
from :
 
$(i)$ Euler's good old
$\Delta^n P(x)=0$, where $\Delta f(x):=f(x+1)-f(x)$, provided
the degree of $P(x)$ is $<n$. (The proof is trivial:
Acting $\Delta$ on any polynomial $Q(x)$,
reduces its degree by $1$.)
 
$(ii)$ The indefinite sum $P(k):=\sum_{k'=1}^{k} p(k')$ of a polynomial
$p(k)$ is a polynomial of degree one higher in $k$. (One way to
prove this trivial fact is to prove it for the basis
$p_i(k):={{i+k} \choose {i}}$, for which
the indefinite sum is $p_{i+1}(k)$, thanks to Pascal's
triangle.)
 
$(iii)$ 
The elementary symmetric functions, $\{e_r\}$, are polynomials in the
power-sum functions $\{p_r\}$. 
This follows from (e.g. Macdonald's {\it Symmetric Functions})
\hfill\break
$\sum_{r=0}^{n} e_r t^r=\prod_{i=1}^n (1+x_i t)=
\exp(\log (\prod_{i=1}^n (1+x_i t)))=
\exp(\sum_{i=1}^n\log  (1+x_i t))= \exp(\sum_{r=1}^{\infty} (-1)^r 
{{p_r} \over {r}} t^r)$.
 
$(iv)$ The factorizations  $k^4-i^4=(k-i)(k+i)(k^2+i^2)$ and
$i^4+4k^4=((i-k)^2+k^2)((i+k)^2+k^2)$, and cancellations in products.
 
Using their method, Almkvist and Granville[AG] concocted many other
non-shaloshable identities.
 
Another `master identity' that one should keep in mind when
trying to prove `strange' identities, is the Lagrange\footnote{$^3$}
{\baselineskip=9pt
\eightrm  \raggedright Also the Lagrange Inversion Formula,
used so beautifully by Ira Gessel and Dennis Stanton, but that's
another story.}
Interpolation
Formula: if $deg P(x) < n$ then,
$$
P(x)= \sum_{k=1}^{n} {{\prod_{{{i=1} \atop {i \neq k}}}^{n} (x-x_i)}
\over {\prod_{{{i=1} \atop {i \neq k}}}^{n} (x_k-x_i)}} P(x_k) \quad.
$$
Recall the trivial proof: both sides are polynomials of degree $<n$
that match at the $n$ values $x_1 , \dots , x_n$. By taking
$P(x)$ to be, for example,
$$
P(x)=\prod_{j=1}^{n-1} (x+y_j),
$$
or more complicated forms, and choosing $x_i$ and $y_j$ in 
countless ways, it is possible to come up with many identities,
hypergeometric or otherwise, that would stump Shalosh.
 
{\bf REFERENCES}
 
[AG] Gert Almkvist and Andrew Granville, {\it Borwein and
Bradley's Ap\'ery-like formulae for $\zeta(4n+3)$}, preprint.
{\tt http://www.math.uga.edu/\Tilde andrew/Postscript/BorBrad.ps}.
 
 
[BB] Jon Borwein and David Bradley, {\it Empirically determined
Ap\'ery-like formulae for $\zeta(4n+3)$}, Experimental Math,
to appear. Downloadable from CECM's Home Page
{\tt http://www.cecm.sfu.ca}.
 
 
\bye
 
 
 
 
 
 
 
