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Suppose that $n$ is a natural number and $b$ is one coprime to it such that $b^n\equiv 1\pmod n$. Does it follow that, if $b^x-x\equiv b^y-y\pmod n$, then $x\equiv y\pmod n$?

This is inspired by the comments on a previous answer I wrote, which got me thinking about fixed points on exponentiation. Toying around in Mathematica, I noticed that $$x\mapsto 9^x-x\pmod{1000}$$ is an injective function. In fact, I noticed that if we replace $9$ by any other number coprime to $1000$, this still holds. Toying around further, I noticed that the list of numbers $n$ other than $1000$ such that all coprime bases $b$ had this property starts the same way as this sequence in OEIS, which is exactly the sequence of $n$ such that $b^n\equiv 1\pmod n$ for $b$ coprime to $n$. I can't prove the lists are equal (but they seem to be), but I can see that $b^n\equiv 1\pmod n$ is a necessary condition for $x\mapsto b^x-x$ to be injective (as it's not even well-defined otherwise).

The only insight I have into the problem is that it clearly holds if $b=1$ or $-1$. The latter can be argued as if $(-1)^n\equiv 1\pmod n$ then $n$ is even and $x\mapsto (-1)^x-x$ maps evens to odds and vice versa and is a linear function with unit ($-1$) coefficient of $x$ over those domains - and hence is bijective. I feel like there may be some generalization of this approach by partitioning $(\mathbb Z/n\mathbb Z,+)$ into $\text{ord}_n(b)$ cosets (where $\text{ord}$ is the multiplicative order of $b$ mod $n$), but I have been unsuccessful in finding it if it exists. To be sure, restricting the map in question to a single such coset yields a bijection between it and another coset (as $b^{x}-x$ and $b^{x+\text{ord}_n(b)}-x-\text{ord}_n(b)$ are equal mod $\text{ord}_n(b)$) - but I can't prove that two distinct cosets do not map to the same place, as this requires considering the map over the larger domain where it is less well-behaved.

I am primarily interested in the answer to this question, regardless of technique, however I would be especially interested in knowing if the methods I sketched can be gainfully extended to a proof.

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Milo Brandt
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1 Answers1

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We can realize the proof sketch in the question as a proof by induction on $n$. The statement holds trivially for $n=1$. Next, choose any $b$ and $n$ such that $b^n\equiv 1\pmod n$. If we let $k$ be the multiplicative order of $b$ mod $n$, we can clearly see that $k$ divides $n$, as $b^x\equiv 1$ only for multiples of $k$, and further that $k<n$, as $k$ is the order of the multiplicative subgroup group generated by $b$, and this must divide the order of the multiplicative group mod $n$, which is less than $n$.

Now suppose that $$b^x-x\equiv b^y-y\pmod n.\tag{1}$$ Notice that this implies the weaker statement that $b^x-x\equiv b^y-y \pmod k$. However, since $b^k\equiv 1\pmod k$ and $k<n$, we can apply the inductive hypothesis and conclude that $x\equiv y\pmod k$. However, this implies $b^x\equiv b^y\pmod n$. Subtracting equation (1) from this yields $x\equiv y\pmod n$, as desired.

Milo Brandt
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