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# 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.

Can you answer this question?

Toolbox:
• $(a+b)6n=nC_0a6n+nC_1a^n-1b+nC_2a^{n-2}b^2+.....nC_ra^{n-r}b^r+....nC_nb^n$
Now $a=a+b-b$
$\Rightarrow b+(a-b)$
$a^n=(b+(a-b))^n=b^n+nC_1a^{n-1}(a-b)+nC_2b^{n-2}(a-b)^2+(a-b)^n$
Or $a^n-b^n=nC_1b^{n-1}(a-b)+nC_2(a-b)^2+....+(a-b)^n=a-b$
$\Rightarrow (a-b)[nC_1b^{n-1}+nC_2b^{n-2}(a-b)+...+(a-b)^{x-1}$]
Thus $(a-b)$ is a factor of $(a^n-b^n)$
Hence proved.
answered Jun 23, 2014