# Let $K$ be a field with characteristic $p>0$ and set $q=p^{n}$ for $n \in \mathbb{N}^*$. Prove $(a+b)^{q}=a^{q}+b^{q}\ \forall a,b \in K$.

abstract-algebrafield-theory

Let $$K$$ be a field with characteristic $$p>0$$ and set $$q=p^{n}$$ for $$n \in \mathbb{N}^*$$. Prove $$(a+b)^{q}=a^{q}+b^{q}\ \forall a,b \in K$$.

I can prove using the coefficient theorem that $$(a+b)^{p}=a^{p}+b^{q}$$, by saying that $$p\mid {p\choose i}$$ for $$0\leq i \leq p-1$$. The problem is we have to prove it for $$q$$ that is not a prime number, which doesn't allow the same reasoning.

Let $$q=np$$ for $$n\in \mathbb{Z}^+$$ and consider the expression $$(a+b)^{np}=\sum_{k=0}^{np}{np \choose k} a^{np-k}b^k$$ and again notice that if $$p\mid {np \choose k}$$ for $$k\in \lbrace 1,2,3,\dots, np-1\rbrace$$ then the claim follows, but it is false in general.
Let $$n=2$$ and $$p=3$$. Then $$3$$ doesn't divide $${6 \choose 3}=20$$.
Probably you want write $$q=p^n$$ for $$\mathbb{Z}^+$$ in which case the assertion is true, and the proof consist of give a easy generalization for $$p\mid {p^n \choose k}$$ for $$k\in \lbrace 1,2,\dots, p^n-1\rbrace$$ which can be done by Lucas Theorem.