Question

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]

Answer 1

In order to prove that (a-b) is a factor of \((a^n - b^n)\), it has to be proved that \(a^n - b^n= k (a^n - b^n)\), where k is some natural number
It can be written that, a= a - b + b
\(∴ a^n (a-b+b)^n =[(a-b)+b]^n\)
\(= C_0^n (a-b)^n +C_1^n (a-b)^{n-1} b+...+ C_{n-1}^n(a-b)b^{n-1} + C_n^n b^n\)
\(=(a-b)^n +c_1^n (a-b)^{n-1} b+...+ C_{n-1}^n (a - b) b^{n-1}+b^n\)
\(⇒ a^n − b^n = (a−b)[(a−b)^{n-1} + C_1^n (a - b)^{n-2} b + ...+ C_{n-1}^n\,b^{n-1}]\)
\(⇒a^n-b^n= k(a-b)\)
where, k =\( [(a − b)^{n-1} + C_1^n (a - b)^{n-2} b+...+C_{n-1}^n\,b^{n-1}]\) is a natural number
This shows that (a - b) is a factor of \((a^n - b^n)\), where n is a positive integer