Intuitive explanation of solutions to a linear diophantine equation












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"Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"



I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...






diophantine-equations intuition linear-diophantine-equations







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    1












    $begingroup$


    "Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"



    I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...






    diophantine-equations intuition linear-diophantine-equations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      "Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"



      I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...






      diophantine-equations intuition linear-diophantine-equations







      share|cite|improve this question









      $endgroup$




      "Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"



      I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...






      diophantine-equations intuition linear-diophantine-equations




      diophantine-equations intuition linear-diophantine-equations






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 7 at 13:33









      stochasticmrfoxstochasticmrfox

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          $begingroup$

          Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.



          Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.



          But, sort of as a dufus, I could insist on exchanging more bills.



          So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$



          So we would still exchange exactly $c$






          share|cite|improve this answer









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            $begingroup$

            Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.



            Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.



            But, sort of as a dufus, I could insist on exchanging more bills.



            So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$



            So we would still exchange exactly $c$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.



              Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.



              But, sort of as a dufus, I could insist on exchanging more bills.



              So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$



              So we would still exchange exactly $c$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.



                Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.



                But, sort of as a dufus, I could insist on exchanging more bills.



                So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$



                So we would still exchange exactly $c$






                share|cite|improve this answer









                $endgroup$



                Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.



                Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.



                But, sort of as a dufus, I could insist on exchanging more bills.



                So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$



                So we would still exchange exactly $c$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 13:57









                JuliusL33tJuliusL33t

                1,3681917




                1,3681917






























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