Using Binomial Theorem to prove divisibility

Question

How to use the binomial Theorem to prove that for all integers $x$ and $y$, $a\mid (a+b)^n-b^n$.

I am stuck after expanding the expression, which is $$ a \mid \sum_{m=0}^n \binom{n}{m}a^{n-m}b^m-b^n $$

Could anyone please give me a hint about how to proceed after this step? Greatly appreciated!

Answer 1

$$(a+b)^n - b^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \ldots + \binom{n}{n-2}a^2b^{n-2} + \binom{n}{n-1}ab^{n-1} + b^n - b^n \\= a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \ldots + \binom{n}{n-2}a^2b^{n-2} + \binom{n}{n-1}ab^{n-1} = a\left(a^{n-1} + \binom{n}{1}a^{n-2}b + \binom{n}{2}a^{n-3}b^2 + \ldots + \binom{n}{n-2}ab^{n-2} + \binom{n}{n-1}b^{n-1}\right)$$

Thus $a \mid (a+b)^n - b^n$.

Answer 2

Hint. In the expansion of $$(a+b)^n,$$ all terms but one are of the form $af(a,b).$ The only odd one is $b^n,$ so that when this is subtracted off the remainder has $a$ as factor.

An obvious way to see this is to use the formula for the difference of two powers.

Answer 3

You have $$\sum_{m=0}^{n}{n\choose m}a^{n-m}b^{m}-b^n$$ $$=\sum_{m=0}^{n-1}{n\choose m}a^{n-m}b^{m}+{n\choose n}a^{n-n}b^{n}-b^n$$ $$=\sum_{m=0}^{n-1}{n\choose m}a^{n-m}b^{m}+b^{n}-b^n=\sum_{m=0}^{n-1}{n\choose m}a^{n-m}b^{m}$$

and $a|a^{n-m}$ for each $0\leq m\leq n-1$ since $n-m\geq 1$.