Let $p>3$ be an odd prime and $a$ be a positive integer, is the following congruence true?


When $a=1$, this is Morley's congruence.

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    $\begingroup$ Granville generalized Morley's result in a different direction. He has shown that for any $m\ge 2$ one has $$(*) \prod_{i=1}^{m-1} \binom{p-1}{\lfloor \frac{ip}{m} \rfloor} \equiv (-1)^{\frac{(p-1)(m-1)}{2}} m^{m(p-1)} \bmod {p^2},$$ which gives a weaker version of Morley' congruence when $m=2$. Moreover, he has shown that $(*)$ holds modulo $p^3$ iff the $p-2$'nd Bernoulli Polynomial vanishes mod $p$ at $\frac{1}{m}, \frac{2}{m}, \cdots, \frac{m-1}{m}$. This way Morley's congruence is recovered at full. See eq. 15 here: dms.umontreal.ca/~andrew/PDF/BinCoeff.pdf . $\endgroup$ – Ofir Gorodetsky Oct 20 '16 at 16:33

The answer is yes. See Corollary 3 of "A congruence involving the quotients of Euler and its applications (I)" by Tianxin Cai (Acta Arithmetica, 2002).

Namely, Cai shows that $$(-1)^{\frac{p-1}{2}} \binom{p^k-1}{\frac{p^k-1}{2}}/\binom{p^{k-1}-1}{\frac{p^{k-1}-1}{2}} \equiv 4^{\phi(p^k)} \bmod {p^{3k}}.$$ In particular, for any $k\ge 1$, $$(-1)^{\frac{p-1}{2}} \binom{p^k-1}{\frac{p^k-1}{2}}/\binom{p^{k-1}-1}{\frac{p^{k-1}-1}{2}} \equiv 4^{\phi(p^k)} \bmod {p^{3}}.$$ By applying the above for all $1 \le k \le a$, he deduces that $$(−1)^{\frac{(p-1)a}{2}} \binom{p^a-1}{\frac{p^a-1}{2}} \equiv 4^{p^a-1} \bmod {p^3}.$$ To see that this is the same as your congruence, note that $(p-1)a \equiv p^a-1 \bmod 4$ by considering two cases: $p \equiv 1 \bmod 4$ and $p \equiv -1 \bmod 4$.


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