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Here is the answer from lecture notes. $(a+b)^p = \sum {{p}\choose{k}}a^kb^{p-k}$ and all the terms divide $p$ except $a^p$ and $b^p$ terms. So reducing (mod p) all terms are zero except the ones above.

So I don't understand how by reducing (mod p) makes those terms $0$? By definition of prime characteristic, $p.1=0$ where $1$ and $0$ are the multiplicative and additive identities. So I am not able to make the connection between the definition and the statement above. Thanks and appreciate a hint.

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  • $\begingroup$ By the definition of prime characteristic $p\cdot1=0$ (not $1^p=0$). $\endgroup$ – Lord Shark the Unknown Feb 10 at 22:13
  • $\begingroup$ That's true, correcting that. $\endgroup$ – manifolded Feb 10 at 22:14
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    $\begingroup$ $\binom{p}{k}$ is a multiple of $p$ if $k\in \{1,\dots, p-1\}.$ $\endgroup$ – mfl Feb 10 at 22:15
  • $\begingroup$ Thanks, I understand that but don't understand how that would make the terms $0$ for $k\in \{1,..p-1\}$? $\endgroup$ – manifolded Feb 10 at 22:17
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    $\begingroup$ $p\cdot 1=0\implies p\cdot x=0,\forall x\in R.$ $\endgroup$ – mfl Feb 10 at 22:18
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$$ {{p}\choose{k}} = \frac{p\cdot(p-1)\cdot\ldots\cdot{(p-k+1)}}{k!} $$ As $p$ is prime and $0 < k < p$, $p$ doesn't divide $k!$, so this expression is a multiple of $p$ and thus it's zero in characteristic $p$.

Therefore, the sum $$ \sum_{k=0}^p {{p}\choose{k}}a^kb^{p-k} = a^p + b^p + \sum_{k=1}^{p-1} {{p}\choose{k}}a^kb^{p-k} = a^p + b^p + \sum_{k=1}^{p-1}0\cdot a^kb^{p-k} = a^p+b^p $$

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Because for all $k$ with $0<k<p$ the binomial coefficient $\binom{p}{k}=\frac{p!}{k!(p-k)!}$ is a multiple of $p$, because the numerator divisible by $p$, but the denominator isn't.

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