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Exercise 5.1 · Q12

Q.If aa and bb are distinct integers, prove that a−ba-b is a factor of an−bna^n-b^n, whenever nn is a positive integer. Hint: write an=(a−b+b)na^n=(a-b+b)^n and expand.

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Expand an=((a−b)+b)na^n=\big((a-b)+b\big)^n by the binomial theorem; the k=0k=0 term is exactly bnb^n, and every other term visibly contains a factor of (a−b)(a-b) — factoring that out proves the divisibility.

Step 1. Write a=(a−b)+ba=(a-b)+b and apply the binomial theorem to ana^n.

an=((a−b)+b)n=∑k=0nnCk (a−b)k bn−k.a^n = \big((a-b)+b\big)^n = \sum_{k=0}^n {}^nC_k\,(a-b)^k\,b^{n-k}.

Step 2. Separate the k=0k=0 term. The k=0k=0 term is nC0(a−b)0bn=bn{}^nC_0(a-b)^0b^n=b^n. So

an−bn=∑k=1nnCk (a−b)k bn−k.a^n - b^n = \sum_{k=1}^n {}^nC_k\,(a-b)^k\,b^{n-k}.

Step 3. Factor (a−b)(a-b) out of every remaining term. Each term in the sum (k≥1k\ge1) carries at least one power of (a−b)(a-b):

an−bn=(a−b)[∑k=1nnCk (a−b)k−1 bn−k].a^n-b^n = (a-b)\left[\sum_{k=1}^n {}^nC_k\,(a-b)^{k-1}\,b^{n-k}\right]. …

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