Binomial sum modulo prime power: Part 2

Let p be a prime number and \omega=\exp(2\pi i/p). Recall that we conjectured in this post that

(*)\qquad\qquad\qquad\displaystyle\sum_{j=0}^{p-1}(1-\omega^j)^n=p\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^{kp}\binom{n}{kp}

is divisible by p^{\lceil n/(p-1)\rceil}. From exercise 2 of this post we know that (p) factors as (p,\omega-1)^{p-1} into prime ideals in the ring \mathbb Z[\omega]. So (\omega-1)^{p-1}\in (p). Therefore the right-hand side of (*) is divisible by p^{\lfloor n/(p-1)\rfloor}. So we are off by at most one factor of p! I believe this approach can be improved upon to account for the extra factor, since we haven’t used any property of the sum.

Furthermore,

\displaystyle\sum_{j=0}^{p-1}\omega^{-rj}(1-\omega^j)^n=p\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}

for any r. So our best result thus far is the following:

Theorem (weaker version of Fleck’s).

\displaystyle\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}\equiv 0\pmod{p^{\lfloor\frac{n}{p-1}\rfloor-1}}.

Goal: Improve the floors to ceilings.

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Filed under Algebra, Number theory

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