How to solve $5=a+2b$, $3=c+2d$ with $ad-bc = pm 1$?












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In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:



1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?



2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?






algebra-precalculus







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    1












    $begingroup$


    In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:



    1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?



    2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?






    algebra-precalculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:



      1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?



      2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?






      algebra-precalculus







      share|cite|improve this question











      $endgroup$




      In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:



      1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?



      2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 7:45







      72D

















      asked Dec 19 '18 at 17:27









      72D72D

      417117




      417117






















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          $begingroup$

          Solve the two equations individually first.



          $a+2b=5$ gives $a=2n+1, b=2-n.$



          $c+2d=3$ gives $c=2m+1, d=1-n.$



          Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.



          EDIT



          I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.






          share|cite|improve this answer











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          • $begingroup$
            $d=1-m$ ........
            $endgroup$
            – lhf
            Dec 21 '18 at 9:16













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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Solve the two equations individually first.



          $a+2b=5$ gives $a=2n+1, b=2-n.$



          $c+2d=3$ gives $c=2m+1, d=1-n.$



          Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.



          EDIT



          I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $d=1-m$ ........
            $endgroup$
            – lhf
            Dec 21 '18 at 9:16


















          2












          $begingroup$

          Solve the two equations individually first.



          $a+2b=5$ gives $a=2n+1, b=2-n.$



          $c+2d=3$ gives $c=2m+1, d=1-n.$



          Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.



          EDIT



          I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $d=1-m$ ........
            $endgroup$
            – lhf
            Dec 21 '18 at 9:16
















          2












          2








          2





          $begingroup$

          Solve the two equations individually first.



          $a+2b=5$ gives $a=2n+1, b=2-n.$



          $c+2d=3$ gives $c=2m+1, d=1-n.$



          Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.



          EDIT



          I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.






          share|cite|improve this answer











          $endgroup$



          Solve the two equations individually first.



          $a+2b=5$ gives $a=2n+1, b=2-n.$



          $c+2d=3$ gives $c=2m+1, d=1-n.$



          Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.



          EDIT



          I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 18:21

























          answered Dec 19 '18 at 18:14









          saulspatzsaulspatz

          16k31331




          16k31331












          • $begingroup$
            $d=1-m$ ........
            $endgroup$
            – lhf
            Dec 21 '18 at 9:16




















          • $begingroup$
            $d=1-m$ ........
            $endgroup$
            – lhf
            Dec 21 '18 at 9:16


















          $begingroup$
          $d=1-m$ ........
          $endgroup$
          – lhf
          Dec 21 '18 at 9:16






          $begingroup$
          $d=1-m$ ........
          $endgroup$
          – lhf
          Dec 21 '18 at 9:16




















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