If $p$ is prime, then $(x+y)^p=x^p+y^p$ holds in any field of characteristic $p$. (proved with the help of the Binomial Theorem) But now I need to prove that this implies that this is true for any $(a+b+c...+n)^p = a^p + b^p + \cdots + n^p$. The paper says I have to prove it via Induction. I want to prove:
$$(a_1+\cdots+a_n)^p=a_1^p+\cdots+a_n^p$$
Thank you for any answers!
On
Let $S(n)$ be the statement that $(a_1+\dots+a_n)^p\equiv a_1^p+\dots+a_n^p$. We know that $S(1)$ is true trivially, and $S(2)$ is the usual Freshman's dream. Now that we have our base cases, we just need to show that $S(n)$ implies $S(n+1)$ for all $n\ge 2$ to complete the inductive proof.
Assume $S(n)$. To prove $S(n+1)$, note $$ \begin{aligned} (a_1+\dots+a_n+a_{n+1})^p &=(\color{blue}{(a_1+\dots+a_n)}+a_{n+1})^p \\&\stackrel{S(2)}= (\color{blue}{a_1+\dots+a_n})^p+(a_{n+1})^p \\&\stackrel{S(n)}= a_1^p+\dots+a_n^p+a_{n+1}^p \end{aligned} $$ We have proven $S(n+1)$, so the proof by induction is complete.
Well, it is obviously true for one term and also for two via the binomial theorem. Now, suppose the assertion holds for $k-1$ ($k\ge3$) terms, i.e.
$$(a_1+\cdots+a_{k-1})^p=a_1^p+\cdots+a_{k-1}^p\,.$$
Then
\begin{align*} (a_1+\cdots+a_{k-2}+a_{k-1}+a_k)^p &=(a_1+\cdots+a_{k-2}+(a_{k-1}+a_k))^p \\ &=a_1^p+\cdots+a_{k-2}^p+(a_{k-1}+a_k)^p \\ &=a_1^p+\cdots+a_{k-2}^p+a_{k-1}^p+a_k^p \end{align*}
where we first use the induction hypothesis and then the case $k=2$.