TITLE: Binomial supercongruences: is there any reason for them?
QUESTION [17 upvotes]: One of the recent questions, in fact
the answer
to it, reminded me about the binomial sequence
$$
a_n=\sum_{k=0}^n{\binom{n}{k}}^2{\binom{n+k}{k}}^2,
\qquad n=0,1,2,\dots,
$$
of the Apéry numbers. The numbers $a_n$ come as
denominators of rational approximations to $\zeta(3)$ in Apéry's
famous proof of the irrationality of the number.
There are many nice properties of the sequence, one of these
is the observation that for primes $p\ge5$,
$$
a_{pn}\equiv a_n\pmod{p^3},
\qquad n=0,1,2,\dots.
$$
The congruence was conjectured for $n=1$ by S. Chowla, J. Cowles and M. Cowles
and proved in the full generality by I. Gessel (1982).
There is probably nothing strange in the congruence
(which belongs, by the way, to the class of supercongruences as it happens
to hold for a power of prime higher than one). But already the classical
binomials behave very similarly: for primes $p\ge5$,
$$
\binom{pm}{pn}\equiv\binom mn\pmod{p^3},
$$
the result due to G.S. Kazandzidis (1968). There are many other examples
of modulo $p^3$ congruences, most of them explicitly or implicitly related
to some modular objects, but that is a different story. My question is:
what are the grounds for the above (very simple) congruence for binomial
coefficients to hold modulo $p^3$? Not modulo $p$ or $p^2$ but $p^3$.
I do not ask you to prove the supercongruence but to indicate a general
mechanism which provides some kind of evidence for it and can be used
in other similar problems.
My motivation rests upon my own research on supercongruences; most of them
are just miracles coming from nowhere...

REPLY [5 votes]: Andrew Granville, in his beautiful article Binomial coefficients modulo prime powers (1997), attributes to Jacobsthal the congruence
$$
\binom{np}{mp} \Big/ \binom{n}{m} \equiv 1 \mod{p^{3+\mathrm{ord}_p\{nm(n-m)\}}} \quad \quad \quad (p \geq 5).
$$
In Chapter 7 of Alain's Course in $p$-Adic Analysis (GTM 198), the same congruence is attributed to Kazandzidis. 
This is a (mild?) generalization of Wolstenholme's theorem, $\binom{np}{mp} \equiv \binom{n}{m} \mod{p^3}$.
As both references explain, the general congruence is best understood via the $p$-adic analog of Stirling's asymptotic formula for the $\Gamma$-function. The $p$-adic logarithm of the left-hand side admits a convergent power series expansion involving the $p$-adic Bernoulli numbers - just as the higher order terms in the usual Stirling formula involve the usual Bernoulli numbers. This is covered by formulas (35) and (38) in Granville's article. When $p \geq 5$, all terms in the expansion are seen to have at least the stated valuation $\delta := 3+\mathrm{ord}_p\{nm(n-m)\}$, and the only term with valuation possibly $\delta$ has the form $B_{p-3} \times p^3 \times nm(n-m) \times u_p$, where $u_p \in \mathbb{Z}_p^{\times}$ is a $p$-adic unit. 
Therefore the congruence holds modulo $p^{4 + \mathrm{ord}_p(nm(n-m))}$ if and only if $p \mid B_{p-3}$. Those are the Wolstenholme primes referred to by Greg Kuperberg in his answer.
[I came to this old question via the "Related" column for the new post A congruence involving binomial coefficients by Richard Stanley, which proposes a different generalization of Wolstenholme's $\binom{2p}{p} \equiv 2 \mod{p^3}$. ]