Hi all, just wondering what the method is to find an integer with order t mod m.
For example, find all integers that have order 11 (mod 45).
I know you can geuss and test, but I'm thinking there must be a faster way.
Thanks.
One last thing:
If $\displaystyle a\in \mathbb{Z}/m\mathbb{Z} $ is a primitive root, then we know that $\displaystyle \mathbb{Z}/m\mathbb{Z} = <a> $ i.e. $\displaystyle \forall \; b\in \mathbb{Z}/m\mathbb{Z}, \; b=a^k $ for some $\displaystyle k\in \mathbb{N} $.
So all one has to do to find $\displaystyle ord_m(b) $ for any $\displaystyle b\in \mathbb{Z}/m\mathbb{Z} $, is find $\displaystyle ord_m(a) $ and use the formula $\displaystyle ord_m(b) = ord_m(a^k)=\frac{ord_m(a)}{(ord_m(a),k)} $.
In summary, knowing the order of a primitive root gives you the order of every number in your group.