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Math Help - Diophantine question

  1. #1
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    Diophantine question

    Hi plz give me a help to find the answer for following

    If a \ne 0, gcd(a,b)=g and (x_0,y_0) is a solution of ax+by=c then show that
    x=x_0+\frac{b}{g}t, y=y_0-\frac{a}{g}t
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  2. #2
    Super Member Bacterius's Avatar
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    What is t ?
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by dhammikai View Post
    Hi plz give me a help to find the answer for following

    If a \ne 0, gcd(a,b)=g and (x_0,y_0) is a solution of ax+by=c then show that
    x=x_0+\frac{b}{g}t, y=y_0-\frac{a}{g}t
    Here we are dealing with integer solution of
    ax+by=c

    Since,
    (x_0,y_0) is a solution
    ax+by=ax_0+by_0
    a(x-x_0)+b(y-y_0)=0
    Since there are infinite integer solution(why?), let us introduce an integer parameter m such that x-x_0=m
    It becomes
    am+b(y-y_0)=0
    y-y_0=\frac{am}{b}
    Now, to ensure that RHS is always an integer, make substitution m=\frac{bt}{g}....where t is new integer parameter.
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  4. #4
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    (1) ax + by = c
    (2) ax_{0} + by_{0} = c is an integer solution.

    Subtracting (2) from (1),
    a\left(x-x_{0}\right)+b\left(y-y_{0}\right)=0.

    Now
    (*) \frac{a}{g}\left(x-x_{0}\right)=-\frac{b}{g}\left(y-y_{0}\right).

    Thus
    \frac{b}{g}\mid \frac{a}{g}\left(x-x_{0}\right). But \left(\frac{b}{g},\frac{a}{g}\right)=1 therefore,

    \frac{b}{g}\mid \left(x-x_{0}\right). Hence it follows that there is a t \in \mathbb{Z} such that

    (3) x-x_{0} = t \frac{b}{g}. Substituting this in (*),

    \frac{a}{g} \cdot t\frac{b}{g} = -\frac{b}{g}\left(y-y_{0}\right). Hence y-y_{0}=-t\frac{a}{g}

    Therefore, from above and (3),
    x = x_{0} + t\frac{b}{g}

    y=y_{0}-t\frac{a}{g}
    QED
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  5. #5
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    OK Thanks, I got the idea for the answer, thanks again
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