Do the following system of equation have a solution in $mathbb{Z}_{p}$












1












$begingroup$


Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.



I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?






linear-algebra systems-of-equations







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    1












    $begingroup$


    Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.



    I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?






    linear-algebra systems-of-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.



      I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?






      linear-algebra systems-of-equations







      share|cite|improve this question











      $endgroup$




      Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.



      I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?






      linear-algebra systems-of-equations




      linear-algebra systems-of-equations






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 10:43









      Sujit Bhattacharyya

      1,409519




      1,409519










      asked Dec 7 '18 at 10:40









      user46697user46697

      215211




      215211






















          1 Answer
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          $begingroup$

          As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both



          but in $pmod3$ or $Z_3,$ there is no inverse of $-3$



          In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$



          Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. Can you please explain how you got 2(mod 7).
            $endgroup$
            – user46697
            Dec 7 '18 at 11:02










          • $begingroup$
            @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 11:04











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          1 Answer
          1






          active

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          active

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          2












          $begingroup$

          As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both



          but in $pmod3$ or $Z_3,$ there is no inverse of $-3$



          In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$



          Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. Can you please explain how you got 2(mod 7).
            $endgroup$
            – user46697
            Dec 7 '18 at 11:02










          • $begingroup$
            @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 11:04
















          2












          $begingroup$

          As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both



          but in $pmod3$ or $Z_3,$ there is no inverse of $-3$



          In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$



          Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. Can you please explain how you got 2(mod 7).
            $endgroup$
            – user46697
            Dec 7 '18 at 11:02










          • $begingroup$
            @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 11:04














          2












          2








          2





          $begingroup$

          As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both



          but in $pmod3$ or $Z_3,$ there is no inverse of $-3$



          In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$



          Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?






          share|cite|improve this answer











          $endgroup$



          As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both



          but in $pmod3$ or $Z_3,$ there is no inverse of $-3$



          In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$



          Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 11:03

























          answered Dec 7 '18 at 10:43









          lab bhattacharjeelab bhattacharjee

          226k15158275




          226k15158275












          • $begingroup$
            Thank you. Can you please explain how you got 2(mod 7).
            $endgroup$
            – user46697
            Dec 7 '18 at 11:02










          • $begingroup$
            @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 11:04


















          • $begingroup$
            Thank you. Can you please explain how you got 2(mod 7).
            $endgroup$
            – user46697
            Dec 7 '18 at 11:02










          • $begingroup$
            @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 11:04
















          $begingroup$
          Thank you. Can you please explain how you got 2(mod 7).
          $endgroup$
          – user46697
          Dec 7 '18 at 11:02




          $begingroup$
          Thank you. Can you please explain how you got 2(mod 7).
          $endgroup$
          – user46697
          Dec 7 '18 at 11:02












          $begingroup$
          @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
          $endgroup$
          – lab bhattacharjee
          Dec 7 '18 at 11:04




          $begingroup$
          @user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
          $endgroup$
          – lab bhattacharjee
          Dec 7 '18 at 11:04


















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