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Question
If a and b are distinct integers, prove that a − b is a factor of an − bn, whenever n is a positive integer. [Hint: write an = (a − b + b)n and expaand]
Solution
a = a − b + b
So, an = [a − b +b]n
= [(a − b) + b]n
= nC0 (a − b)n + nC1 (a − b)n−1b1 + nC2 (a − b)n−2b2 + ....... + nCn−1 (a − b)bn−1 + nCn (bn)
⇒ an − bn = (a − b)n + nC1 (a − b)n−1b + nC2 (a − b)n−2b2 + ....... + nCn−1 (a − b)bn−1
= (a − b) [(a − b)n−1 + nC1 (a − b)n−2b + nC2 (a − b)n−3b2 + ...... + nCn−1 bn−1]
= (a – b)[an integer]
⇒ an – bn is divisible by (a – b)
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