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Ax+by=c proof

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    For, a, b, c[tex]\in[/tex]integers and D[tex]\in[/tex]integers -{0}, if a and b are divisible by d, and c is not divisible by d then the equation ax+by=c has no integral solution for x and y.


    2. Relevant equations



    3. The attempt at a solution
    ax+by=c
    a/dx+b/dx=c
    1/d[ax+by]=c
    ax+by=cd
     
  2. jcsd
  3. Nov 17, 2008 #2

    HallsofIvy

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    Where did the d in a/d come from? You can't just "stick" it into part of the equation.

    Saying "d divides a" means a= dn for some integer n. Saying "d divides b" means b= dm for some integer m. Replace a and b in your equation by that and see what happens.
     
  4. Nov 17, 2008 #3
    dnx+dmx=c
    d(nx+mx)=c
     
  5. Nov 18, 2008 #4
    I don't know if that was the correct way to do it and then maybe say since we don't have a dpc then there is no integral solution?
     
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