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Suppose that $p$ is a prime. Suppose further that $h$ and $k$ are non-negative integers such that $h + k = p − 1$.

I want to prove that $h!k! + (−1)^h \equiv 0 \pmod{p}$

My first thought is that by Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$, and $h!k!$ divides $(p-1)!$ (definition of a binomial). Where would I go from here?

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2 Answers 2

up vote 3 down vote accepted

Use the fact that

$$h! = (-1)^h (p-1)(p-2) \dots (p-h) \mod p$$

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Ah, so h!k! becomes [(-1)^h](p-1)! mod p, which is just -(-1)^h. Thanks so much! –  Johan Feb 27 '11 at 22:59

This is simply a reformulation of the Wilson reflection formula that I mentioned twice yesterday

$$\rm (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p),\ \ for\ p\ prime $$

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