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Math Help - Linear Diophantine Equations

  1. #1
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    Linear Diophantine Equations

    12x+13y=14

    gcd(12,13)=1\rightarrow \ 1|14

    Can the equations x and y be found using the vector <12,16> \ \mbox{and} \ (-1,2) \mbox{?}

    x=-1+12t
    y=2+13t

    Does this work?
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  2. #2
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    I'm not sure I understand your two vectors <12,16> and <-1,2>.

    It is true that (x_0,y_0) = (-1,2) is a solution to the equation 12x + 13y = 14 and that the general solution is of the form (x_0 + a \cdot t, y_0 + b \cdot t) where (a,b) are integer constants, however in this case the constants are not (12,13) but rather (13,12).

    Can you see why?

    Hence the general solution is x = -1 + 13 \cdot t and y = 2 + 12 \cdot t
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  3. #3
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    There aren't two vectors. That is why I said vector not vectors. The vector is denoted <,> and (,) denotes a point.

    You know how if you have a vector and a point you can parametrize it? Can that be done for these types of equations?
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  4. #4
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    It depends on whether or not the point is a solution the equation.

    Regardless of the point used as your initial solution (x_0,y_0), there is only ONE vector v = <v_1,v_2> that can be used to describe the entire set of solutions as (x_0 + v_1, y_0 - v_2)

    Is this what you wanted?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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