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Let $p$ be a prime number. Prove that in the field $\mathbb Z_p$ the following is true $(x+y)^p= x^p + y^p$

My idea to do this is to apply the binomial theorem, namely:
$$(x+y)^p = x^p + px^{p-1}y+ {p \choose 2}x^{p-2}y^2+ \dots+ {p \choose {p-2}}x^2y^{p-2}+ pxy^{p-1} + y^p$$ Now, I have made an observation (I do not know if this is true):
$$(\forall n)((n \in \mathbb Z_5 \land n \ne 0 \land n \ne p) \Rightarrow (\exists q\in \mathbb Z_p)({p\choose n}= pq) $$ And so all of these elements corresponding to $n \in \{1,2, \dots, p-2, p-1 \}$ will turn into zeros, and the only remaining elements will be $x^p$ and $y^p$. Is my reasoning good? If so, should I have used induction here?

Aemilius
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2 Answers2

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The crucial point is that $p$ divides $p \choose k$ (in $\mathbb{Z}$) for any prime $p$ and $k=1,\dots,p-1.$

This follows for example from $$ k {p \choose k} = p {p-1 \choose k-1} $$ by noting that $p$ and $k$ are relative prime.

Reiner Martin
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To elaborate on Reiner Martin's answer, $p$ divides $\displaystyle \binom{p}{k}$ for $0 < k < p$, which can be seen from its formula:

$$\binom{p}{k} = \frac{p!}{k!(p-k)!}$$

We know (due to combinatoric argument) that when $0 < k < p$ that $\displaystyle \binom{p}{k}$ is integer. However, since $p$ is prime and $k < p \wedge p-k < p$ we have that neither $k!$ nor $(p-k)!$ have $p$ as factor. Thus we can rewrite the above formula as such, where the fraction is still integer:

$$\binom{p}{k} = p\cdot\frac{(p-1)!}{k!(p-k)!}$$

Which clearly proves the conjecture.

orlp
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