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(From Math Challenge II Number Theory packet)

Given that $a,b,n$ are positive integers. Assume that for any positive integer $k\neq b, (k-b)\mid(k^n-a)$, the which of the following must be true?

(A) $a>b^n$

(B) $a<b^n$

(C) $a=b^n$

(D) It depends.

I'm not sure how to approach this problem. Can someone please explain?

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  • $\begingroup$ C must be true. Dont know how to argue about the others though. C follows from difference of powers identity. $\endgroup$ – Sandeep Silwal May 14 '14 at 4:27
  • $\begingroup$ Post an answer. It's multiple choice, so you only have to prove one of them is correct. $\endgroup$ – Jason Chen May 14 '14 at 4:28
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Since $k - b$ divides $k^n-b^n$ for any $k$, it also divides $$ (k^n - a) - (k^n - b^n) = b^n - a $$ for any $k \neq b$. Therefore $b^n - a$ has infinitely many distinct divisors, and so $b^n - a = 0$, which is (C).

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  • $\begingroup$ $k-b$ divides $(k^n-a)$, but how did you get $(k^n-b^n)$? $\endgroup$ – Jason Chen May 14 '14 at 4:49
  • $\begingroup$ $k^n-b^n = (k-b)(k^{n-1} + k^{n-2}b + \cdots + kb^{n-2} + b^{n-1})$ $\endgroup$ – Cardboard Box May 14 '14 at 5:09
  • $\begingroup$ Another way of seeing this: since $b$ is a root of the polynomial $X^n - b^n$, this polynomial is divisible by $X-b$. $\endgroup$ – Cardboard Box May 14 '14 at 5:10

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