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.

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- Mar 18th 2010, 01:55 PMseven.jMultiplicative Order
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. - Mar 18th 2010, 02:22 PMchiph588@
- Mar 18th 2010, 02:30 PMchiph588@
- Mar 18th 2010, 02:42 PMchiph588@
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.