Find the orders, if.. Suppose that $ord_ma=6$.  Find the orders $ord_m(a^2)$ and $ord_m(a^5)$.  Explain why your answers are correct.
What I know:
For $ord_ma=6$ I believe this is saying $a^6 \equiv 1 (mod m)$.  Assuming this is correct, from here I would only be guessing, because the book / teachers lecture notes have hardly covered this, it seems we were kind of expected to know this before taking this course.
Anyway, my guess would be that the order for $ord_m(a^2)=12$, because $6*2=12$, but like I said, nothing but a guess.
Thanks,
Chloe :)
 A: Let me give a hint for the first part - you will probably be able to get the other part then.
If $ord_ma=6$, then we know that $a^6\equiv 1 \pmod{m}$ and also that no smaller power of $a$ gives $1 \pmod{m}$.
The question is asking for the smallest power of $a^2$ which gives $1 \pmod{m}$, and we certainly have $(a^2)^3 \equiv 1 \pmod{m}$, so that 3 looks like it should be the answer - but we need to check that no smaller power of $a^2$ is $\equiv 1 \pmod{m}$. If there actually were a smaller power of $a^2$ which is $\equiv 1 \pmod{m}$, what could you deduce?
A: $\def\ord{\operatorname{ord}}$$\ord_m a = 6$ means two things:


*

*As you stated correctly, $a^6 \equiv 1\pmod m$

*6 has to be the smallest positive number that fulfills this, i. e. we must have $a^i \not\equiv 1 \pmod m$ for $1 \le i \le 5$.


To compute $\ord_m a^2$ we compute it's powers: 


*

*$(a^2)^1 = a^2$. We now (as $2 < 6$) that $a^2 \not\equiv 1\pmod m$.

*$(a^2)^2 = a^4$, here also $a^4 \not\equiv 1 \pmod m$.

*$(a^3)^2 = a^6$, here $a^6 \equiv 1 \pmod 6$.


As 3 was the smallest number for which we obtained 1, $\ord_m a^2 = 3$. I'm sure, you can do $a^5$ on your own.
