Prove that $(n-m) \mid (n^r - m^r)$

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.

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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 $n^{r+1}-m^{r+1}=n(n^r-m^r)+m^r(n-m)$.

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+1 for more methods of proof. – Callum Rogers Jan 17 '11 at 0:21

This is a simple telescopic sum trick: $n^{r} - m^{r} = (n-m) \sum_{k=1}^{r} n^{r-k}m^{k-1} = (n - m)(n^{r-1} + n^{r-2}m + \cdots +nm^{r-2} + m^{r-1}).$

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Very nice, was trying some very similar to this but not with enough terms. Thanks. – Callum Rogers Jan 17 '11 at 0:20

HINT $\$ modulo $\rm\ n-m\$ one has $\rm\ n\equiv m\ \Rightarrow\ n^r \:\equiv\: m^r\$ by the congruence product rule.

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Nice, this seems to me the most elegant and simple answer! +1 – wildildildlife Apr 30 '11 at 20:46

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.

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HINT $\$ For $\rm\ n\to n+m\$ it is $\rm\ n\ |\ (n+m)^r - m^r\$ which is obvious by the binomial theorem.

Note $\$ The above proof arises from the following trivial symmetry-inspired proof of the

Factor Theorem $\rm\ \ \ \ x - m \ |\ 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 the shift $\rm\ x\to x+m\:.\$ QED

The proof exploits an innate symmetry - the problem is invariant under shift automorphisms. Hence, with luck, we can apply these symmetries to reduce the problem to a much simpler one.

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HINT $\$ Put $\rm\ P(x) = x^r,\ \ x = n\$ in $\rm\ x-m\ |\ P(x)-P(m)\$ in $\rm\ \mathbb Z[x]\$ (the Factor Theorem)

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