Question

If $a$ and $b$ are distinct integers, prove that $a-b$ is factor of $a^n-b^n$, whenever $n$ is a positive integer.

Answer 1

For $n=1,a^n-b^n=a-b$

Let for $n=k,a^k-b^k=\lambda (a-b)$

for $n=k+1,a^n-b^n=a^{k+1}-b^{k+1}$

$=a.a^k-b.b^k=a[b^k+\lambda (a-b\left)\right]-b^{k+1}$

$=a{b}^k-b^{k+1}+\lambda a(a-b)$

$=b^k(a-b)+\lambda a(a-b)$

$=(a-b)(b^k+\lambda a)$

$={\lambda }^{\prime }(a-b)$

$\Rightarrow$ By induction $a^n-b^n$ is always divisible by $a-b$