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Number Theory, with a proof discussed in class (not homework)
Could someone give me a hint?- Worded.Mouse
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- Forum: Calculus and Beyond Homework Help
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Number Theory, with a proof discussed in class (not homework)
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[SIZE="3"]Z Let m=m'd. Then ax=mk+1=d(m'k)+1- Worded.Mouse
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- Class Homework Number theory Proof Theory
- Replies: 1
- Forum: Calculus and Beyond Homework Help