If $a$ and $b$ are distinct integers, prove that $a-b$ is a factor of $a^{n}-b^{n}$, whenever $n$ is a positive integer.
[Hint: write $a^{n}=(a-b+b)^{n}$ and expand]
In order to prove that $(a-b)$ is a factor of $\left(a^{n}-b^{n}\right)$, it has to be proved that
$a^{n}-b^{n}=k(a-b)$, where $k$ is some natural number
It can be written that, a = a – b + b
$\therefore a^{n}=(a-b+b)^{n}=[(a-b)+b]^{n}$
$={ }^{n} \mathrm{C}_{0}(a-b)^{n}+{ }^{n} \mathrm{C}_{1}(a-b)^{n-1} b+\ldots+{ }^{n} \mathrm{C}_{n-1}(a-b) b^{n-1}+{ }^{n} \mathrm{C}_{n} b^{n}$
$=(a-b)^{n}+{ }^{n} \mathrm{C}_{1}(a-b)^{n-1} b+\ldots+{ }^{n} \mathrm{C}_{n-1}(a-b) b^{n-1}+b^{n}$
$\Rightarrow a^{n}-b^{n}=(a-b)\left[(a-b)^{n-1}+{ }^{n} \mathrm{C}_{1}(a-b)^{n-2} b+\ldots+{ }^{n} \mathrm{C}_{n-1} b^{n-1}\right]$
$\Rightarrow a^{n}-b^{n}=k(a-b)$
where, $k=\left[(a-b)^{n-1}+{ }^{n} C_{1}(a-b)^{n-2} b+\ldots+{ }^{n} C_{n-1} b^{n-1}\right]$ is a natural number
This shows that $(a-b)$ is a factor of $\left(a^{n}-b^{n}\right)$, where $n$ is a positive integer.