Use the binomial theorem to prove if $m\mid b - a$, then $m \mid b^n - a^n$.

Question

I'm trying to prove that if $m\mid b - a$, then $m \mid b^n - a^n$. I have done it several ways so far, including through induction and through the application of theorems regarding congruence (i.e. if $a\equiv b\pmod m$ then $a^n\equiv b^n\pmod n$).

I am now faced with trying to prove this using the binomial theorem, i.e. $$(a +b)^n = \sum_{k=0}^n{n \choose k} a^k b^{n-k}$$ The hint I am given is to write $b$ as $b=c+a$ where $c=b-a$, and to expand $(c+a)^n$ using the binomial theorem.

Where I'm running into problems is trying to relate a sum of polynomials to divisibility, aka how am I to relate $${n \choose 0} b^n +{n \choose 1} a c^{n-1} +...+{n \choose n-1} a^{n-1} c + {n \choose n} a^n$$ to $m\mid b - a$? My current thinking is to somehow factor out $(b - a)$ from the string of polynomials, but I can't seem to get that cleanly.

Any direction/help is appreciated!

Answer 1

We have \begin{align*} b^n - a^n &= \bigl(a + (b-a)\bigr)^n - a^n\\ &= \sum_{k=0}^n \binom nk a^k(b-a)^{n-k} - a^n\\ &= \sum_{k=0}^{n-1} \binom nk a^k(b-a)^{n-k} \end{align*} In the last sum, every summand contains at least one factor $(b-a)$, hence this sum is a multiple of $m$. Therefore $m \mid b^n - a^n$.

Answer 2

$$b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+\ldots+ba^{n-2}+a^{n-1})$$

Answer 3

Hint Write

$$b=a+mk$$

and use the binomial theorem to calculate $b^n$.

What is $b^n-a^n$ and why is this divisible by $m$?

Answer 4

Note that $((b-a)+a)^n-a^n=b^n-a^n$ and use the binomial theorem on the first term.