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I have a question regarding this proof:

If $a$ and $m$ are coprime, then $a^{t} ≡ 1 $ (mod m) for some t, $1 ≤ t < m$. Since $a$ and $m$ are coprime, $m$ does not divide as for any $s$, and so the $m$ numbers $1,a,a^{2}, . . . ,a^{m-1}$ all belong to the $m−1$ congruence classes other than the congruence class of $[0]$.

So two of the numbers must be in the same congruence class: that is, there exist numbers $s$ and $t$ with $s ≥ 0$ and $0 <t ≤ m−1$ so that $a^{s} ≡ a^{t+s}$ (mod m). Now since $a$ and $m$ are coprime, we can cancel the common factor as from both sides of the congruence to get $1 ≡ a^{t}$ (mod m).

I can not really justify this line: "$a^{s} ≡ a^{t+s}$ (mod m)"

My guess is that for $a^{t+s}$ I can obtain all possible residues, so $a^{t+s}$ gives residues in one "periode" and since $a^{t}$ does not reside in this periode but in another, $a^{s} ≡ a^{t+s}$ is gonna be true at some point. But for that I have to proof that if I take powers of $a:1,a,a^{2}, . . . ,a^{m-1}$, there is a repetition after a maximum of m-1 steps. And I am not sure how to show that, if that is even the reason behind the congruence "$a^{s} ≡ a^{t+s}$ (mod m)".

Thanks for your help

Banause
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  • Didn't you just post this question the other day? https://math.stackexchange.com/questions/4173640/modular-arithmetic-proof-not-clear – Gerry Myerson Jun 19 '21 at 12:27
  • You are right. Same proof but different question. Since the question was closed I do not know a good way to combine them. – Banause Jun 19 '21 at 12:44
  • By pigeonhole there are $,s',s,$ with $,a^{\large s'}!\equiv a^{\large s},,$ wlog $,s'>s.,$ Let $,t := s'-s,$ so $,s' = s+t.,$ Then $,t\ge 1,$ by $,s'>s,$ and $,t\le m-1,$ since that's the maximum possible index difference $,s'-s,$ (which occurs when $,s' = m-1,$ is largest, and $,s = 0,$ is smallest) – Bill Dubuque Jun 19 '21 at 20:18
  • We applied pigeonhole to the map $,f(k) := a^{k}\bmod m.,$ It maps $,m,$ pigeons $\in{0,1,\dots,m!-!1},$ into $,m-1,$ holes = all $\rm\color{#0a0}{nonzero}$ residues or congruence classes $!\bmod m,,$ where $,\color{#0a0}{f(k)\not\equiv 0},,$ by $,\gcd(a,m)=1\Rightarrow \gcd(\color{#c00}{a^k},m)=1,,$ so $,\color{#0a0}{m\nmid a^k},,$ where we used that coprimes to $,m,$ (invertibles) are closed under $\rm\color{#c00}{products}$. $\ \ $ – Bill Dubuque Jun 19 '21 at 23:25
  • You can't combine them, but in the interests of transparency you can link them, that is, include in each a link to the other. Also, you can read the information given to you about why the other question was closed, and make improvements to get it reopened. – Gerry Myerson Jun 20 '21 at 00:48

1 Answers1

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It's essentially just using the pigeon hole principle, there are m "objects" that is our numbers

$1,a,a^2,...,a^{m-1}$

And they have to be "placed" in m-1 "boxes" that is the residue

$1,2,...,m-1$

So there will be at least two that are in the same "box" ie have the same residue

$a^t=a^{t+s} $(mod m)

And then canceling a^t gives us the result

Mehul
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  • Thanks for your answer. I understand this principle. But how I can be sure that the two objects that reside in the same class, does not reside in the right-hand side: $a^{t+s}$ (mod m). In other words. Let's say m = 7, then there are these possible residue classes: {1,2,3,4,5,6}. What if we have for any s the residue 1 on the left-hand-side, and for all t's we get following residue on the right-handside: {2,3,4,5,6} since 2 of the powers have the same residue of 2. – Banause Jun 19 '21 at 11:25
  • I think I can answer my own question. Since $1,a,a^{2},...,a^{m−1}$ share at least one class. We compare each a with all other powers. For example: m = 7 and $1,a,a^{2},...,a^{m−1}$ are {1,2,3,4,3,5,6}. Then we compare 1 with {2,3,4,3,5,6}, 2 with {1,3,4,3,5,6} and 3 with {1,2,4,3,5,6} and so on. And in this case for 3 we find the congruence. – Banause Jun 19 '21 at 11:54
  • I didn't understand what u mean by make sure that they don't reside on the rhs, because they are distinct, one will have larger exponent and one smaller, so assign t for the smaller a t+s for the larger and then it's done – Mehul Jun 19 '21 at 12:21
  • Sorry I wrote $a^{t}≡a^{t+s}$ instead of $a^{s}≡a^{t+s}$. Maybe that is the reason we do not really understand each other. I edited it – Banause Jun 19 '21 at 12:42