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Modular Congruence and GCD Proof

  1. Feb 21, 2015 #1
    1. The problem statement, all variables and given/known data
    If a≡b(mod n) and d=gcd(a.n), prove that d=gcd(b,n).



    2. Relevant equations
    If a≡b(mod n) → n|(a-b) → a-b =nk, for some k∈ℤ → a=nk+b
    If d=gcd(a.n) → d=ax+ny
    gcd(b,n)=d ↔ d|b and d|n, and if c|b and c|n, then c ≤ a.


    3. The attempt at a solution
    Since a=nk+b and d=ax+ny, then
    d=(nk+b)x+ny
    d=nkx+bx+ny
    d=bx+n(kx+y)
    d=bx+nw, where w=kx+y and w∈ℤ.

    I have gotten this far. Originally I thought this was enough to prove d was the gcd of b and n. But now I see that I need to show d|b, d|n, and that if there is a c such that c|b and c|n, then c ≤ d.

    Help!
     
  2. jcsd
  3. Feb 21, 2015 #2

    Dick

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    Homework Helper

    It's pretty clear that if d|n and d|a then d|b, right? Show that. Now if there is a smaller c that also divides b and n, then it also must divide a and n, right? Show that. Put the two together.
     
    Last edited: Feb 21, 2015
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