Using the binomial theorem to prove for all integers $a$ and $b$, $a \big| \big((a + b)^n - b^n\big)$

Question

How can binomial theorem be used to prove for all integers $a$ and $b$, $a \big| \big((a + b)^n - b^n\big)$?

So I'm kinda lost on this question. Where do I start? What is the solution?

Answer 1

Expand $(a+b)^n-b^n$ using the Binomial theorem: $$ (a+b)^n-b^n=-b^n+\sum_{k=0}^n {n \choose k}a^kb^{n-k} = -b^n+ b^n+a \times (\sum_{k=1}^n {n \choose k}a^{k-1}b^{n-k}) $$ So the RHS is $a\times(\sum_{k=1}^n {n \choose k}a^{k-1}b^{n-k})$ which is divisible by $a$.

Answer 2

All terms in the expansion but the first contain a positive power of $a$. So if you subtract the first term $b^n$ from the expansion, what remains has $a$ in every term, so you can factor out an $a$. Since $a$, $b$, and all of the coefficients are integers, what remains is also an integer.

$$(a+b)^n = \sum_{k=0}^n {n\choose k} a^k b^{n-k}$$ so since the first term is ${n\choose 0}a^0b^{n-0}=b^n$, we have $$(a+b)^n-b^n= \sum_{k=1}^n {n\choose k} a^k b^{n-k} $$

$$= a\sum_{k=1}^n {n\choose k} a^{k-1} b^{n-k} $$