If a and b are distinct integers, prove that a – b is a factor of an – bn,

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.