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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 :)

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    As a hint, if $a^x \equiv 1$ then $\mathrm{ord}(a)\mid x$2012-10-08

2 Answers 2