1

Let $p$ be a prime number. For any $1 ≤ r ≤ p − 1$, prove that $${p \choose r} ≡ 0 \pmod p$$

I'm thinking that it suffices to show $p$ divides ${p \choose r}$. So then:

$$\begin{align} p\ |\ {p \choose r} &= p\ |\ {p!\over r!(p - r)!} \\ &= p\ |\ p!{1\over r!(p - r)!} \\ &= p\ |\ p(p - 1)!{1\over r!(p - r)!} \\ &= p\ |\ p{(p - 1)!\over r!(p - r)!} \end{align}$$

Thus $\displaystyle p\ |\ {p \choose r}$.

But this seems a little too simple to me. Is this really it, or have I missed something? Thanks for any help!

Ali Caglayan
  • 5,726
Guest
  • 81
  • What is ${ p\choose r}?$ Please write it between $$$'s. – mfl Oct 19 '14 at 17:18
  • 1
    If you knew that $\frac{(p-1)!}{r!(p-r)!}$ was an integer, you'd be fine (though your $=$ signs don't really mean equality; say "if and only if" instead). But you don't know that (unless you prove it). – Micah Oct 19 '14 at 17:31

1 Answers1

3

$$\binom pr=\frac{p(p-1)\cdots(p-r+1)}{r!}$$

Now using Proof that binomial coefficient is a natural number, $\displaystyle(r!)\mid \underbrace{p(p-1)\cdots(p-r+1)}$

But $(r!,p)=1$ for $1\le r\le p-1\implies (r!)\mid \underbrace{(p-1)\cdots(p-r+1)}$

Community
  • 1
lab bhattacharjee
  • 274,582