Exercise 2.1.45* (Binomial congruences 2)

Question

Show that if $p$ is prime then $\binom{p}{k} \equiv 0 \pmod p$ for $1\leq k \leq p-1$.

Richard Ganaye · Accepted Answer

\begin{proof}

If $1\leq k \leq p-1$, then for every $i \in [\![1,k]\!]$, $i \wedge p = 1$, hence $k! \wedge p = 1$.

Similarly, for every $j \in [\![1,p-k]\!]$, $j \wedge p = 1$, hence $(p-k)! \wedge p = 1$, so $k!(p-k)! \wedge p = 1$.

Since 
$$p \mid p! = k!(p-k)! \binom{p}{k},\qquad k!(p-k)! \wedge p = 1,$$
we obtain $p \mid \binom{p}{k},$ so
$$\forall k \in [\![1,p-1]\!],\ \binom{p}{k}\equiv 0 \pmod p.$$
\end{proof}

Exercise 2.1.45* (Binomial congruences 2)

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Exercise 2.1.45* (Binomial congruences 2)

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