Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that if $p$ is an odd prime and $k$ is an integer satisfying $1\leq k \leq p-1$,then the binomial coefficient
$$\binom{p-1}{k} \equiv (-1)^k\pmod p$$

I have tried basic things like expanding the left hand side to $\frac{(p-1)(p-2).........(p-k)}{k!}$ but couldn't get far enough.

share|cite|improve this question
up vote 8 down vote accepted

Hint: $(p-1)(p-2)\cdots(p-k)\equiv(-1)(-2)\cdots(-k)$ because $p\equiv 0$.

share|cite|improve this answer
Got it .Thanks for the hint . – Saurabh Jun 8 '12 at 17:15

Hint: For the first few odd primes $p$, write down the $p$th row of Pascal's Triangle modulo $p$; you'll notice a pattern that should be straightforward to prove. Now use the relation $$\binom{p-1}{k-1}+\binom{p-1}{k}=\binom{p}{k}$$ and an induction argument.

share|cite|improve this answer
I couldn't get the hint.I noticed $\binom{p}{k} \equiv 0 \pmod p$ but then I couldn't proceed. – Saurabh Jun 8 '12 at 17:35
We know that $\binom{p-1}{0}=1$, hence $\binom{p-1}{0}\equiv (-1)^0\bmod p$. If $\binom{p-1}{k}\equiv (-1)^k\bmod p$, then $$\binom{p-1}{k}+\binom{p-1}{k+1}=\binom{p}{k+1}$$ and therefore $$(-1)^k+\binom{p-1}{k+1}\equiv 0\bmod p$$ hence $\binom{p-1}{k+1}\equiv (-1)^{k+1}\bmod p$. – Zev Chonoles Jun 8 '12 at 17:38
@SaurabhHota: It is complete. – Babak S. Jun 8 '12 at 17:38

Recall that we have $(x + 1)^p \equiv x^p + 1 \bmod p$. The ring $\mathbb{F}_p[x]$ is an integral domain, so we can divide and it follows that $$(x + 1)^{p-1} \equiv \frac{x^p + 1}{x + 1} \equiv x^{p-1} - x^{p-2} \pm ... - x + 1 \bmod p.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.