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Let $a, b, n \in \mathbb N$. Prove that if $b - k$ divides $a - k^n$ for all $k \in \mathbb N$ such that $k \ne b$, then $a = b^n$.

My approach inspired by this problem and this answer goes somewhat like:

In order to take care of the exponential $k^n$, let's use the Fermat Little Theorem, so let's pick a prime $p$ and consider $k = a^p$.

Then taking $$b - k | a - k^n$$ modulo $p$ implies: $$b - a^p | a - a^{pn} \mod p$$ $$b - a | a - a^n \mod p$$

And here I got stuck. I understand that we have much freedom in our choice of $p$, so we can take it as large as we want, which would also imply that it's not a divisor of $b - a$ nor $a - a^n$, but still I don't understand how I can deduce the equality of values from their divisibility modulo $p$.

Andrey Surovtsev
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2 Answers2

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Let $a=b^n+c$,

then $a-k^n=b^n-k^n+c$ is divisible by $b-k$ for arbitrary integer k not equal to $b$.

Since $b^n-k^n$ is divisible by $b-k$,

therefore $c$ is divisible by $b-k$.

Since $k$ is arbitrary integer not equal to $b$, $b-k$ is an arbitrary nonzero integer.

If $c \neq 0$, the conclusion that $c$ is divisible by any nonzero integer is impossible because $c$ is not divisible by $2c$.

Thus $c$ must be $0$ and $a=b^n$.

Li Kwok Keung
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Choose $k>b+|b^n-a|$. Then notice that since $b-k \mid a-b^n$ and $b-k \mid b^n-k^n$, $b-k \mid b^n-a$, which implies that either $|b-k|\leq|b^n-a|$ or $b^n-a=0$.

But $b-k<-|b^n-a|\leq 0$, which implies that $|b-k|>|b^n-a|$. So $b^n-a=0$ and $a=b^n$.

Piotr H
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