If $ord^a_m = k$ and $a^s \equiv a^t \pmod{m}$ then show $\gcd(k,s) = \gcd(k,t)$.

Question

If $ord^a_m = k$ and $a^s \equiv a^t \pmod{m}$ then show $$\gcd(k,s) = \gcd(k,t).$$

My attempt: Let $s \le t$. So we have

$$a^{t-s}\equiv 1 \pmod{m}$$

So by definition of order we have $k | t-s$ and hence $s\equiv t \pmod{k}$.

Can you help me?

Answer 1

Well, you are almost there. If $d$ is a common divisor of $k,t$ then $d|t$ and $d|(t-s)$, and so $d|s$. So in this case $d$ is also a common divisor of $k,s$. And in the same way, any common divisor of $k,s$ is a common divisor of $k,t$. In particular, $\gcd(k,t)=\gcd(k,s)$.