Prove that $a^m+b^m=(a+b)^m$

Question

Let $F$ be a field such that $char F=p\neq 0$, let $n>0$ and $m=p^n$ prove that $a^m+b^m=(a+b)^m$ for all $a,b\in F$

Attempt:

Since $a,b\in F$ then $a^m=(1\cdot a)^m=1^ma^m=0\cdot a^m=0$

and $b^m=(1\cdot b)^m=1^mb^m=0\cdot b^m=0\implies a^m+b^m=\color{red}0$

$(a+b)^m=[1\cdot (a+b)]^m=(1)^m(a+b)^m=0\cdot (a+b)^m=\color{red}0$

Relatet, but not the same

$1^m = 1$, not $0$. I think the best way to prove this is by induction on $n$, the exponent in $m= p^n$. The base case, $n=1$, should be known to you as the freshman's dream. The induction step shouldn't be too hard — Maxime Ramzi, May 10 '17 at 10:55
$1^m\neq 0$ in a ring with characteristic $p$: $m\cdot 1=0$. This is just a trivial repeated application of the last question you asked — rschwieb, May 10 '17 at 10:55

score 0 · Accepted Answer · answered May 10 '17 at 10:56

0

Use induction on $n$. Suppose $a^m' + b^m' =(a+b)^m'$ then taking $p$th power and using binomial theorem on LHS we deduce the same holds for $pm$. For $m=p$ this is obvious since $p| pCi$ for all $0<i<p$

answered May 10 '17 at 10:56

Prove that $a^m+b^m=(a+b)^m$

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