# Number Theory, need help with a proof discussed in class (not homework)

1. Oct 14, 2012

### Worded.Mouse

Prove that ordda | ordma, when d|m.

Some conditions are 1 ≤ d, 1 ≤ m, and gcd(a,d)=1.

What I have so far:

let x=ordma, which gives us ax$\equiv$ 1 (mod m) $\Rightarrow$ ax=mk+1 for some k$\in$Z

Let m=m'd. Then ax=mk+1=d(m'k)+1

2. Oct 14, 2012

### Worded.Mouse

Could someone give me a hint?