Prove $n-m \mid n^r - m^r\,$ [Factor Theorem, monomial case] In respect to a larger proof I need to prove that $(n-m) \mid (n^r - m^r) $ (where $\mid$ means divides, i.e., $a \mid b$ means that $b$ modulus $a$ = $0$). I have played around with this for a while now but to no avail and would be grateful if you could help me.
 A: If $r$ is a positive integer, simply notice that $x^r - y^r = (x-y)(x^{r-1} + x^{r-2}y + \cdots + xy^{r-2} + y^{r-1})$ as polynomials.
A: Hint $\ $ For $\rm\ n\to n\!+\!m\ $ it is $\rm\ n\ |\ (n\!+\!m)^{\large r} - m^{\large r}\ $ which is clear, e.g. via Binomial Theorem.
Note $\ $ The above proof arises from the following trivial symmetry-inspired proof of the
Polynomial Factor Theorem $\rm\ \ \ x - m \mid  P(x) - P(m)\ $ in $\rm\: R[x],\ $ for every ring $\rm\:R.$
Proof $\ $ Clear if $\rm\ m = 0\!:\ \ x\ |\ P(x)-P(0).\: $ Else reduce to $\rm\ m = 0\ $ by a shift $\rm\ x\to x\!+\!m.$
The proof exploits an innate symmetry - the problem is invariant under shift automorphisms. Hence, with luck, exploiting the innate symmetry reduces the problem to a much simpler one.

More generally the following is true, as proved here.
Theorem $\ $ TFAE for a polynomial $\,f\in R[x],\,$ and $\,a\in R\,$ a commutative ring.
$(0)\ \ \ f = (x\!-\!a)q + r\ $ for some $\,q\in R[x],\ r\in R\ \ \ $ [Monic Linear Division Algorithm]
$(1)\ \ \ f\bmod x\!-\!a = f(a)\ \ \ \ \ \ \ $ [Polynomial Remainder Theorem]
$(2)\ \ \ f(a) = 0\iff x\!-\!a\mid f\ \ \ $ [Polynomial Factor Theorem]
A: Hint $\ $ Put $\rm\ f(x) = x^r,\ \ x = n\ $ in $\rm\ x-m\ |\ f(x)-f(m)\ $ in $\rm\ \mathbb Z[x]\ $ (Polynomial Factor Theorem)
A: There are several ways of proving this. For example, one can use the binomial theorem $(a+b)^r=\sum_{k=1}^r\binom{k}ra^kb^{n-k}$ with $a=n-m$ and $b=m$. (The change of variables from $n,m$ to $a,b$ is in itself a good idea in problems like this, since it tends to reduce the size of some of the expressions.)
Another way is to use the identity $$n^r-m^r=(n-m)\sum_{k=0}^{r-1}n^{r-1-k}m^k=(n-m)(n^{r-1}+n^{r-2}m+\dots+nm^{r-2}+m^{r-1}).$$
Both of these identities are proved by induction on $r$.
Or, one can do directly an inductive argument: The result is clear if $r=0$, since $n^r-m^r=0$. Given the result for $r$, prove it for $r+1$ using, for example, that $n^{r+1}-m^{r+1}=n(n^r-m^r)+m^r(n-m)$.
A: Hint $\bmod\, n-m\!:\,\ \color{#c00}{n\equiv m}\ \Rightarrow\ \color{#c00}n^r \:\equiv\: \color{#c00}m^r\ $ by Congruence Power Rule (iterated Product Rule)
A: This is a simple telescopic sum trick:
\[
n^{r} - m^{r} = (n-m) \sum_{k=1}^{r} n^{r-k}m^{k-1} = (n - m)(n^{r-1} + n^{r-2}m + \cdots +nm^{r-2} + m^{r-1}).
\]
