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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 expand]
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Solution
In order to prove that (a – b) is a factor of (an – bn), it has to be proved that an – bn = k (a – b), where k is some natural number
It can be written that, a = a – b + b
∴ an = (a - b + b)n = [(a - b) + b]n
= nC0 (a - b)n + nC1 (a - b)n - 1 b + ... + nCn- 1 (a - b)bn - 1 + nCnbn
= (a - b)n + nC1 (a - b)n - 1 + b + ... + nCn - 1 (a - b) bn - 1+ bn
= an - bn = (a - b)[(a - b)n - 1+nC1(a - b)n - 2 b + ... + nCn - 1 bn - 1]
= an - bn = k (a - b)
where, k = [(a - b)n - 1 + nC1(a - b)n - 2 b + ... + nCn - 1bn - 1] is a natural number
This shows that (a - b) is a factor of (an - bn), where n is a positive integer.
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