$a^s \equiv a^t \pmod n$ when $s, t$ are limited by ord(a,n)

Question

Suppose we have $a, n \in \mathbb{N}$ with gcd($a, n$) = $1$ and the multiplicative order of $a$ called $k$, such that $a^k \equiv 1 \pmod n$. The definition of multiplicative order states that

there must be two powers, say $s$ and $t$ and $s > t$, such that $a^s \equiv a^t \pmod n$ https://en.wikipedia.org/wiki/Multiplicative_order

This doesn't apply when $s$ and $t$ are smaller than $k$. My question is, why is that true for all $1 \leq t < s < k$

$a^t \pmod n \neq a^s \pmod n$

I know that it has something to do with the limitation of $s, t$, so that it results in $a^t$ and $a^s$ never having the same remainder when divided by $n$.

$a^{s-t}\equiv 1\pmod n$ – Roddy MacPhee Jul 16 '21 at 17:43 — Roddy MacPhee, Jul 16 '21 at 17:43
See the basic Theorem in the linked dupe. – Bill Dubuque Jul 16 '21 at 20:50 — Bill Dubuque, Jul 16 '21 at 20:50

score 1 · Accepted Answer · answered Jul 16 '21 at 17:48

Suppose on the contrary. $a^t \equiv a^s \pmod{n} \implies n| a^t(a^{s-t}-1)$ Now since $(n,a)=1$, we must have that $n| a^{s-t}-1$ or $a^{s-t} \equiv 1 \pmod{n}$, by the definition of order, $s-t \ge k$ but according to the given information $s-t <k-1<k$. contradiction

$a^s \equiv a^t \pmod n$ when $s, t$ are limited by ord(a,n)

1 Answers1