Competition problem — Number theory

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Number theory Difficulty 6.1 National Olympiad Prove it Soviet Union

Problem:

The natural numbers $a$ , $b$ , $n$ are such that for every natural number $k$ not equal to $b$ , $b - k$ divides $a - k^n$ . Prove that $a = b^n$ .

Solution

Solution:

We have $k^n - a = b^n - a$ (mod $b - k$ ). Hence $b^n - a = 0$ (mod $b - k$ ) for every $k$ not equal to $b$ . But if $b^n$ does not equal $a$ , then by taking $k - b > b^n - a$ we could render the equation false.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.