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Thread: Diophantine question

  1. #1
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    Diophantine question

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

    If $\displaystyle a \ne 0$, gcd(a,b)=g and $\displaystyle (x_0,y_0)$ is a solution of $\displaystyle ax+by=c$ then show that
    $\displaystyle x=x_0+\frac{b}{g}t$, $\displaystyle y=y_0-\frac{a}{g}t$
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  2. #2
    Super Member Bacterius's Avatar
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    What is $\displaystyle t$ ?
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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 $\displaystyle a \ne 0$, gcd(a,b)=g and $\displaystyle (x_0,y_0)$ is a solution of $\displaystyle ax+by=c$ then show that
    $\displaystyle x=x_0+\frac{b}{g}t$, $\displaystyle y=y_0-\frac{a}{g}t$
    Here we are dealing with integer solution of
    $\displaystyle ax+by=c$

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

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

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

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

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

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

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

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

    $\displaystyle y=y_{0}-t\frac{a}{g}$
    $\displaystyle QED$
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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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