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I will prove that $(x+y)^{p^k} = x^{p^k} + y^{p^k}$ for $p$ prime and $k\geq 1$ an integer in $\mathbb{F}_p$. In other words, I will show that $$(x+y)^{p^k} \equiv x^{p^k}+y^{p^k} (\text{mod } p)$$

Isaac
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  • what have you tried? this is a pretty standard question. it may be closed without more information in the question itself from you – kodlu Aug 28 '22 at 01:10
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    Isaac, this question has already been asked a couple of times on this site. Try to use the search function to find one of those. And, of course, welcome on math.stackexchange :) – Severin Schraven Aug 28 '22 at 01:12
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    For example here https://math.stackexchange.com/questions/1819552/proving-alpha-betapn-alphapn-betapn-in-a-finite-field?noredirect=1&lq=1 – Severin Schraven Aug 28 '22 at 01:14

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We have that $$(x+y)^{p^k} = \sum_{n=0}^{p^k}\binom{p^k}{n}x^{p^k-n}y^n $$ $$= x^{p^k}+\binom{p^k}{1}x^{p^k-1}y+\binom{p^k}{2}x^{p^k-2}y^2+\ldots + \binom{p^k}{p^k-2}x^2y^{p^k-2}+\binom{p^k}{p^k-1}xy^{p^k-1}+y^{p^k}$$ So it suffices to show that $p|\binom{p^k}{n}$ for $n\neq 0,p^k$. We have that $$\binom{p^k}{n} = \frac{p^k!}{n!(p^k-n)!} = \frac{p^k(p^k-1)!}{n!(p^k-n)!}= p \cdot \frac{p^{k-1} (p^k-1)!}{n!(p^k-n)!}$$ So we just need to show that $n!(p^k-n)!|p^{k-1}(p^k-1)!$. Well, we know that $n!(p^k-n)!|p \cdot p^{k-1}(p^k-1)! = p^k!$. Since $n!(p^k-n)!$ does not divide $p$, it follows that $n!(p^k-n)!|p^{k-1}(p^k-1)!$.

Isaac
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