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Prove that if $k,a,p$ are fixed positive integers, $n$ is a positive integer, $p$ is prime, $gcd(a,p)=1$, then the set {$na^{n}+1$ $mod$ $p^{k}$} is a complete residue system modulo $p^{k}$.

I found this as a part of a question on a high school olympiad mock test. The original problem is:

Prove that for any positive integers $m,a$, there is an infinite amount of positive integers $n$ such that $m|d(na^{n}+1)$, where $d(x)$ denotes the number of positive integer factors of $x$. (not original wording)

If the complete residue system is proved, then for any $k$, there exists an infinite amount of $n$ such that $p^{k}|na^{n}+1$, in which for every $p^{k}(p-1)$ integer values of n, there are ${p-1}$ that satisfy $p^{k}|na^{n}+1$. Similarly, for each $p^{k+1}(p-1)$ integer values of n, there are $p-1$ that satisfy $p^{k+1}|na^{n}+1$. Notice that there are less values of n that satisfy $p^{k+1}|na^{n}+1$. Thus, there is an infinite amount of integers $n$ that satisfy $p^k||na^n+1$. Therefore, $(k+1)|d(na^n+1)$. let $k=m-1$, and we have arrived at the desired result.

Bill Dubuque
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Antonyliao
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  • Per Greg's note, a complete residue system means you have exactly $p^k$ integers to match, which we likely do not have. What we have is that every residue class will appear $\quad$ As such, there is a fallacy in your final paragraph. We cannot guarantee that each term will appear with some frequency, or even infinitely. For example, fix a prime $p$, then the set {primes mod p} has every residue, but clearly there is only a single residue of 0. – Calvin Lin Nov 27 '24 at 18:27
  • Please don't change the question after answers have been posted. Ask a new question if need be. – Bill Dubuque Nov 28 '24 at 09:14

2 Answers2

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First, a complete residue system modulo $p^k$ is a set of exactly $p^k$ integers that intersects each residue class exactly once. So I wouldn't say that this set is a complete residue system, but rather that it intersects every residue class.

It suffices to show that for every $t\pmod{p^k}$, there exists $n$ such that $na^n\equiv t\pmod{p^k}$. I'll show this for $k=1$ and $k=2$, and I hope the procedure is clear to extend the result to all $k$.

Let's first prove the $k=1$ case. Given $t\pmod p$, choose $m=(p-t)(p-1)$. Then $a^m \equiv 1\pmod p$ by Fermat's little theorem, so $ma^m \equiv m \equiv (-t)(-1) \equiv t\pmod p$.

Now let's consider the case $k=2$. Given $t\pmod{p^2}$, there exists an integer $m$ such that $ma^m \equiv t\pmod p$ by the above argument; so we know that $ma^m \equiv t+ps\pmod{p^2}$ for some integer $s$. Now consider $n=jp(p-1)+m$ for a parameter $j$. We have $a^n\equiv a^m\pmod{p^2}$ by Euler's theorem, so $na^n \equiv na^m \equiv (jp(p-1)+m)a^m \equiv (-jp)a^m + ma^m = (-ja^m+s)p + t\pmod{p^2}$. Therefore it suffices to choose $j$ such that $-ja^m+s \equiv 0\pmod p$, that is, choose $j\equiv a^{p-1-m}s\pmod p$.

Greg Martin
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Below is a simple inductive proof of the type of argument hinted in Greg's answer.

Lemma $ $ If prime $\,p\nmid a\,$ then $\, na^n\equiv b\pmod{\!p^k}\,$ has a root $\,n,\,$ for all $\,b\in\Bbb Z,\ k\in\Bbb N$.

Proof $\ $ Induct on $k.\,$ Base case $\,k\!=\!0\!:\ na^n\equiv b\pmod{\!p^0}$ holds for any $\,n,\,$ e.g. $\,n=0$.

Induct: assume true for $k\,$ so $\,\color{#0a0}{na^n = b \!+\! ip^k}.\,$ Let $N = n\! +\! j(p\!-\!1)p^k,\ \color{#c00}{j \equiv \color{#0a0}i/a^n}\!\pmod{\!p}$

hence $\!\bmod p^{k+1}\!:\ N\color{#0af}{a^{N}}\!\equiv (n \!+\! j(p\!-\!1)p^k)\color{#0af}{a^n}$ $\equiv \color{#0a0}{na^n}\!-jp^k a^n\equiv \color{#0a0}b+ \color{#c00}{(\color{#0a0}i-ja^n})p^k\equiv b$

where above $\,\color{#0af}{a^N\!\equiv a^n}\,$ by $\,N\equiv n\pmod{ \!\underbrace{\phi(p^{k+1})}_{\large (p-1)\:\!p^{\large k}}},\ p\nmid a\,$ & $ $ Euler $\phi$ mod order reduction.

Bill Dubuque
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