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Homework Help: Number Theory, need help with a proof discussed in class (not homework)

  1. Oct 14, 2012 #1
    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[itex]\equiv[/itex] 1 (mod m) [itex]\Rightarrow[/itex] ax=mk+1 for some k[itex]\in[/itex]Z

    Let m=m'd. Then ax=mk+1=d(m'k)+1
  2. jcsd
  3. Oct 14, 2012 #2
    Could someone give me a hint?
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