Suppose the following 2 congruences hold \begin{equation*} (x + \alpha)^n \equiv x^n + \alpha \hspace{4mm} (\text{mod} \hspace{2mm} x^r - 1, \hspace{2mm} p) \\ (x + \alpha)^p \equiv x^p + \alpha \hspace{4mm} (\text{mod} \hspace{2mm} x^r - 1, \hspace{2mm} p) \end{equation*}
I am told that this implies the congruence $$ (x + \alpha)^\frac{n}{p} \equiv x^\frac{n}{p} + \alpha \hspace{4mm} (\text{mod} \hspace{2mm} x^r - 1, \hspace{2mm} p) $$
Can any one please explain to me why this is?
Do you mean $\mod {(x^r-1,p)}$? I assume this.
The only possibilities for $(x^r-1,p)$ are $1$ and $p$.
►$(x^r-1,p)=p$
Because of the well known property $(x+y)^p\equiv x+y\pmod p$ we have
$$(x+a)^{\frac np}\equiv x^{\frac np}+a\pmod p\iff(x+a)^n\equiv(x^{\frac np}+a)^p\equiv x^n+a$$ Therefore only the first condition implies the condition to be proven.
►$(x^r-1,p)=1$
This case is trivial in the sense that all pair of integers are equivalent modulo $1$.