Let $R$ be a commutative ring with unity , of prime characteristic $p$. If $n>1$ is an integer such that $(a+b)^n=a^n+ b^n, \forall a,b \in R$ i.e. the map $f : R \to R$ , given by $f(a)=a^n, \forall a\in R$ is a ring homomorphism , then is it true that $n$ is a power of $p$ ?
Considering the copy of $\mathbb F_p$ that embeds in $R$, since $r^n=r, \forall r \in \mathbb F_p$, so $ p-1 |n-1$. But I am unable to conclude anything else.
The comment of Georges shows that the claim is not true for $p=2$. But what about for odd $p$ ?
Please help.
Consolidating the examples from the comments given by Georges Elencwajg and Max . . .
Let $p$ be a prime, and let $n$ be a positive integer.
Claims:
$\;\;\;\;(1)\;\;\;x^n = x$ for all $x\in F_p$ if and only if $(p-1){\,\mid\,}(n-1)$.
$\;\;\;\;(2)\;\;\;(a+b)^n = a^n + b^n$ for all $a,b\in F_p$ if and only if $(p-1){\,\mid\,}(n-1)$.
In particular, from claim $(2)$, it follows that the identity $(a+b)^n = a^n + b^n$ does not force $n$ to be a power of $p$.
Proof of claim $(1)$:
First suppose $(p-1){\,\mid\,}(n-1)$.
Then we can write $n-1 = k(p-1)$, for some positive integer $k$.
Let $x\in F_p$.
If $x=0$, then $x^n = 0$, so $x^n = x$.
If $x \ne 0$, then $x^{p-1}=1$, hence \begin{align*} x^n &= x^{n-1}x\\[4pt] &=x^{k(p-1)}x\\[4pt] &= \left(x^{p-1}\right)^kx\\[4pt] &=(1)^kx\\[4pt] &=x\\[4pt] \end{align*} Conversely, suppose $x^n = x$ for all $x\in F_p$.
Then $x^{n-1}=1$, for $x \in (F_p)^*$.
But the multiplicative group $(F_p)^*$ is cyclic of order $p-1$.
If $a$ is a cyclic generator of $(F_p)^*$, then since the order of $a$ is $p-1$, and $a^{n-1}=1$, it follows that $(p-1){\,\mid\,}(n-1)$.
This completes the proof of claim $(1)$.
Proof of claim $(2)$:
First suppose $(p-1){\,\mid\,}(n-1)$,
By claim $(1)$, we have $x^n=x$ for all $x\in F_p$, hence $(a+b)^n = a + b = a^n + b^n$.
Conversely, suppose $(a+b)^n = a^n + b^n$ for all $a,b\in F_p$.
It follows that for any positive integer $k$, the equality $$(x_1 + \cdots + x_k)^n = x_1^n + \cdots + x_k^n$$ holds for all $x_1,...,x_k\in F_p$.
Letting $x_1 = \cdots = x_k = 1$, we get the identity $k^n = k$, where $k$ is now interpreted as an element of $F_p$.
It follows that $x^n = x$ for all $x\in F_p$, hence by claim $(1)$, we get $(p-1){\,\mid\,}(n-1)$.
This completes the proof of claim $(2)$.