Let $p$ be a prime. How do you show that the binomial coefficients $\binom{p}{i}$ are divisible by $p$ for $1\leq i\leq p-1$? And how does this result in the congruency $(x+y)^p\equiv x^p+y^p\pmod p$ for $x,y\in\mathbb{Z}$?
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1The answer to this question ishould be in the first lines of this question. – May 20 '15 at 13:36
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@G.Sassatelli You are right, but what is the answer to the second question? – May 20 '15 at 13:38
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Binomial theorem – Christoph May 20 '15 at 13:39
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@Christoph can you show me how to use that in this case? – May 20 '15 at 13:53
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For the first part, we have $$\dbinom{p}i = \dfrac{\dbinom{p-1}{i-1} \cdot p}i$$ where $i \in \{2,3,\ldots,p-1\}$. Recall that if $a \mid (bc)$, if $\gcd(a,b) = 1$, then $a \mid c$. Hence, this means $i \mid \dbinom{p-1}{i-1}$, i.e., $\dbinom{p-1}{i-1} = k i$, where $k \in \mathbb{Z}^+$, This in turn gives us that $$\dbinom{p}i = kp \implies p \mid \dbinom{p}i$$
For the second part, we have $$(x+y)^p = x^p + y^p + \sum_{k=1}^{p-1} \dbinom{p}kx^ky^{p-k}$$ From the previous part, we have that $p$ divides $\dbinom{p}k$. This means $$\sum_{k=1}^{p-1} \dbinom{p}kx^ky^{p-k} \equiv 0\pmod{p}$$ Hence, we have that $$(x+y)^p \equiv x^p + y^p \pmod{p}$$
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