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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

Hint $\ $ Wilson's theorem implies that any complete system of representatives of nonzero remainders mod $\,p\,$ has product $\equiv -1.\,$ In particular this is true for any sequence of $\,p\,$ consecutive integers, after removing its $\rm\color{#c00}{multiple}$ of $\,p.\,$ Your special case is the sequence $\, -h,\,-h\!+\!1,\ldots,-1,\require{cancel}\color{#c00}{\cancel{0,}} 1,2,\ldots, k,\,$ whose product is $\,(-1)^h h!\,k!\equiv -1.\ $ QED

Remark $\ $ This is slight reformulation of the Wilson reflection formula mentioned yesterday

$$ k! = (p\!-\!1\!-\!h)! \equiv \frac{(-1)^{h+1}}{h!}\!\!\pmod p,\ \text{ for $\,p\,$ prime} $$

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