$17 | 29x + 18y implies 17 | 23x + 9y$












3












$begingroup$


So, basically the title.
Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
Plus, does then $17 | 11x + 8y$ ?



Can you please explain what's the idea for doing these type of problems .. ?






elementary-number-theory divisibility







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    So, basically the title.
    Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
    Plus, does then $17 | 11x + 8y$ ?



    Can you please explain what's the idea for doing these type of problems .. ?






    elementary-number-theory divisibility







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      So, basically the title.
      Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
      Plus, does then $17 | 11x + 8y$ ?



      Can you please explain what's the idea for doing these type of problems .. ?






      elementary-number-theory divisibility







      share|cite|improve this question









      $endgroup$




      So, basically the title.
      Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
      Plus, does then $17 | 11x + 8y$ ?



      Can you please explain what's the idea for doing these type of problems .. ?






      elementary-number-theory divisibility




      elementary-number-theory divisibility






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 19 '18 at 12:13







      user626177





























          4 Answers
          4






          active

          oldest

          votes


















          2












          $begingroup$

          For the first part:



          Let us first reduce it mod 17. Hence



          $$29x+18y equiv 12x+y pmod{17}$$
          $$23x+9y equiv 6x+9y pmod{17}$$



          since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
            $endgroup$
            – user626177
            Dec 19 '18 at 12:43






          • 1




            $begingroup$
            So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
            $endgroup$
            – user626177
            Dec 19 '18 at 12:44






          • 1




            $begingroup$
            @someone, it amounts to Gaussian elimination mod $17$.
            $endgroup$
            – lhf
            Dec 19 '18 at 12:45








          • 1




            $begingroup$
            Got it, thought it was harder. Thanks, again :)
            $endgroup$
            – user626177
            Dec 19 '18 at 12:52






          • 2




            $begingroup$
            @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
            $endgroup$
            – Bill Dubuque
            Dec 19 '18 at 17:06



















          2












          $begingroup$

          For the first part :



          $17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$



          For the second part :



          $17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$



          On the other hand :



          $17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            $left.begin{align}
            !!bmod 17!: 0, equiv &, 18y + 29x\
            equiv &, y+12x \
            iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
            end{align}right} $             If so then $ left{ begin{align}
            &, 23x + 9,color{#c00}y\
            equiv & , 6x + 9(color{#c00}{5x})\
            equiv &, 51x,equiv, 0
            end{align}right} $             and $ left{ begin{align}
            &, 11x + 8,color{#c00}y\
            equiv &, 11x + 8(color{#c00}{5x})\
            equiv &, 51x,equiv, 0
            end{align}right}$






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              Clear and simple!
              $endgroup$
              – user626177
              Dec 19 '18 at 17:08



















            1












            $begingroup$

            Here is the ansatz:




            If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.




            This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.



            This argument actually shows




            $17 mid 29x + 18y iff 17 mid 23x + 9y$




            Here is an explanation.



            Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
            Its determinant is $-153$, which is $0$ mod $17$.
            Therefore, its two rows are linearly dependent mod $17$.
            In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)



            For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Adapted from math.stackexchange.com/a/2947271/589
              $endgroup$
              – lhf
              Dec 19 '18 at 12:18






            • 1




              $begingroup$
              Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
              $endgroup$
              – user626177
              Dec 19 '18 at 12:23










            • $begingroup$
              @someone, just compute $9(29x + 18y)$ mod $17$.
              $endgroup$
              – lhf
              Dec 19 '18 at 12:25










            • $begingroup$
              Oh, but whats the trick with 9 ?
              $endgroup$
              – user626177
              Dec 19 '18 at 12:26










            • $begingroup$
              @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
              $endgroup$
              – Shaun
              Dec 19 '18 at 12:29











            Your Answer





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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            For the first part:



            Let us first reduce it mod 17. Hence



            $$29x+18y equiv 12x+y pmod{17}$$
            $$23x+9y equiv 6x+9y pmod{17}$$



            since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
              $endgroup$
              – user626177
              Dec 19 '18 at 12:43






            • 1




              $begingroup$
              So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
              $endgroup$
              – user626177
              Dec 19 '18 at 12:44






            • 1




              $begingroup$
              @someone, it amounts to Gaussian elimination mod $17$.
              $endgroup$
              – lhf
              Dec 19 '18 at 12:45








            • 1




              $begingroup$
              Got it, thought it was harder. Thanks, again :)
              $endgroup$
              – user626177
              Dec 19 '18 at 12:52






            • 2




              $begingroup$
              @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 19 '18 at 17:06
















            2












            $begingroup$

            For the first part:



            Let us first reduce it mod 17. Hence



            $$29x+18y equiv 12x+y pmod{17}$$
            $$23x+9y equiv 6x+9y pmod{17}$$



            since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
              $endgroup$
              – user626177
              Dec 19 '18 at 12:43






            • 1




              $begingroup$
              So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
              $endgroup$
              – user626177
              Dec 19 '18 at 12:44






            • 1




              $begingroup$
              @someone, it amounts to Gaussian elimination mod $17$.
              $endgroup$
              – lhf
              Dec 19 '18 at 12:45








            • 1




              $begingroup$
              Got it, thought it was harder. Thanks, again :)
              $endgroup$
              – user626177
              Dec 19 '18 at 12:52






            • 2




              $begingroup$
              @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 19 '18 at 17:06














            2












            2








            2





            $begingroup$

            For the first part:



            Let us first reduce it mod 17. Hence



            $$29x+18y equiv 12x+y pmod{17}$$
            $$23x+9y equiv 6x+9y pmod{17}$$



            since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?






            share|cite|improve this answer











            $endgroup$



            For the first part:



            Let us first reduce it mod 17. Hence



            $$29x+18y equiv 12x+y pmod{17}$$
            $$23x+9y equiv 6x+9y pmod{17}$$



            since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 19 '18 at 17:17

























            answered Dec 19 '18 at 12:37









            Maged SaeedMaged Saeed

            8771417




            8771417








            • 1




              $begingroup$
              Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
              $endgroup$
              – user626177
              Dec 19 '18 at 12:43






            • 1




              $begingroup$
              So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
              $endgroup$
              – user626177
              Dec 19 '18 at 12:44






            • 1




              $begingroup$
              @someone, it amounts to Gaussian elimination mod $17$.
              $endgroup$
              – lhf
              Dec 19 '18 at 12:45








            • 1




              $begingroup$
              Got it, thought it was harder. Thanks, again :)
              $endgroup$
              – user626177
              Dec 19 '18 at 12:52






            • 2




              $begingroup$
              @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 19 '18 at 17:06














            • 1




              $begingroup$
              Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
              $endgroup$
              – user626177
              Dec 19 '18 at 12:43






            • 1




              $begingroup$
              So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
              $endgroup$
              – user626177
              Dec 19 '18 at 12:44






            • 1




              $begingroup$
              @someone, it amounts to Gaussian elimination mod $17$.
              $endgroup$
              – lhf
              Dec 19 '18 at 12:45








            • 1




              $begingroup$
              Got it, thought it was harder. Thanks, again :)
              $endgroup$
              – user626177
              Dec 19 '18 at 12:52






            • 2




              $begingroup$
              @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 19 '18 at 17:06








            1




            1




            $begingroup$
            Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
            $endgroup$
            – user626177
            Dec 19 '18 at 12:43




            $begingroup$
            Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
            $endgroup$
            – user626177
            Dec 19 '18 at 12:43




            1




            1




            $begingroup$
            So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
            $endgroup$
            – user626177
            Dec 19 '18 at 12:44




            $begingroup$
            So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
            $endgroup$
            – user626177
            Dec 19 '18 at 12:44




            1




            1




            $begingroup$
            @someone, it amounts to Gaussian elimination mod $17$.
            $endgroup$
            – lhf
            Dec 19 '18 at 12:45






            $begingroup$
            @someone, it amounts to Gaussian elimination mod $17$.
            $endgroup$
            – lhf
            Dec 19 '18 at 12:45






            1




            1




            $begingroup$
            Got it, thought it was harder. Thanks, again :)
            $endgroup$
            – user626177
            Dec 19 '18 at 12:52




            $begingroup$
            Got it, thought it was harder. Thanks, again :)
            $endgroup$
            – user626177
            Dec 19 '18 at 12:52




            2




            2




            $begingroup$
            @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
            $endgroup$
            – Bill Dubuque
            Dec 19 '18 at 17:06




            $begingroup$
            @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
            $endgroup$
            – Bill Dubuque
            Dec 19 '18 at 17:06











            2












            $begingroup$

            For the first part :



            $17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$



            For the second part :



            $17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$



            On the other hand :



            $17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              For the first part :



              $17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$



              For the second part :



              $17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$



              On the other hand :



              $17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                For the first part :



                $17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$



                For the second part :



                $17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$



                On the other hand :



                $17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$






                share|cite|improve this answer











                $endgroup$



                For the first part :



                $17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$



                For the second part :



                $17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$



                On the other hand :



                $17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 19 '18 at 13:01

























                answered Dec 19 '18 at 12:34









                MatkoMatko

                864




                864























                    2












                    $begingroup$

                    $left.begin{align}
                    !!bmod 17!: 0, equiv &, 18y + 29x\
                    equiv &, y+12x \
                    iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
                    end{align}right} $                     If so then $ left{ begin{align}
                    &, 23x + 9,color{#c00}y\
                    equiv & , 6x + 9(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right} $                     and $ left{ begin{align}
                    &, 11x + 8,color{#c00}y\
                    equiv &, 11x + 8(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right}$






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Clear and simple!
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 17:08
















                    2












                    $begingroup$

                    $left.begin{align}
                    !!bmod 17!: 0, equiv &, 18y + 29x\
                    equiv &, y+12x \
                    iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
                    end{align}right} $                     If so then $ left{ begin{align}
                    &, 23x + 9,color{#c00}y\
                    equiv & , 6x + 9(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right} $                     and $ left{ begin{align}
                    &, 11x + 8,color{#c00}y\
                    equiv &, 11x + 8(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right}$






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Clear and simple!
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 17:08














                    2












                    2








                    2





                    $begingroup$

                    $left.begin{align}
                    !!bmod 17!: 0, equiv &, 18y + 29x\
                    equiv &, y+12x \
                    iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
                    end{align}right} $                     If so then $ left{ begin{align}
                    &, 23x + 9,color{#c00}y\
                    equiv & , 6x + 9(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right} $                     and $ left{ begin{align}
                    &, 11x + 8,color{#c00}y\
                    equiv &, 11x + 8(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right}$






                    share|cite|improve this answer









                    $endgroup$



                    $left.begin{align}
                    !!bmod 17!: 0, equiv &, 18y + 29x\
                    equiv &, y+12x \
                    iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
                    end{align}right} $                     If so then $ left{ begin{align}
                    &, 23x + 9,color{#c00}y\
                    equiv & , 6x + 9(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right} $                     and $ left{ begin{align}
                    &, 11x + 8,color{#c00}y\
                    equiv &, 11x + 8(color{#c00}{5x})\
                    equiv &, 51x,equiv, 0
                    end{align}right}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 19 '18 at 17:01









                    Bill DubuqueBill Dubuque

                    212k29195648




                    212k29195648








                    • 2




                      $begingroup$
                      Clear and simple!
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 17:08














                    • 2




                      $begingroup$
                      Clear and simple!
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 17:08








                    2




                    2




                    $begingroup$
                    Clear and simple!
                    $endgroup$
                    – user626177
                    Dec 19 '18 at 17:08




                    $begingroup$
                    Clear and simple!
                    $endgroup$
                    – user626177
                    Dec 19 '18 at 17:08











                    1












                    $begingroup$

                    Here is the ansatz:




                    If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.




                    This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.



                    This argument actually shows




                    $17 mid 29x + 18y iff 17 mid 23x + 9y$




                    Here is an explanation.



                    Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
                    Its determinant is $-153$, which is $0$ mod $17$.
                    Therefore, its two rows are linearly dependent mod $17$.
                    In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)



                    For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Adapted from math.stackexchange.com/a/2947271/589
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:18






                    • 1




                      $begingroup$
                      Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:23










                    • $begingroup$
                      @someone, just compute $9(29x + 18y)$ mod $17$.
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:25










                    • $begingroup$
                      Oh, but whats the trick with 9 ?
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:26










                    • $begingroup$
                      @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
                      $endgroup$
                      – Shaun
                      Dec 19 '18 at 12:29
















                    1












                    $begingroup$

                    Here is the ansatz:




                    If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.




                    This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.



                    This argument actually shows




                    $17 mid 29x + 18y iff 17 mid 23x + 9y$




                    Here is an explanation.



                    Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
                    Its determinant is $-153$, which is $0$ mod $17$.
                    Therefore, its two rows are linearly dependent mod $17$.
                    In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)



                    For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Adapted from math.stackexchange.com/a/2947271/589
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:18






                    • 1




                      $begingroup$
                      Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:23










                    • $begingroup$
                      @someone, just compute $9(29x + 18y)$ mod $17$.
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:25










                    • $begingroup$
                      Oh, but whats the trick with 9 ?
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:26










                    • $begingroup$
                      @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
                      $endgroup$
                      – Shaun
                      Dec 19 '18 at 12:29














                    1












                    1








                    1





                    $begingroup$

                    Here is the ansatz:




                    If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.




                    This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.



                    This argument actually shows




                    $17 mid 29x + 18y iff 17 mid 23x + 9y$




                    Here is an explanation.



                    Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
                    Its determinant is $-153$, which is $0$ mod $17$.
                    Therefore, its two rows are linearly dependent mod $17$.
                    In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)



                    For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.






                    share|cite|improve this answer











                    $endgroup$



                    Here is the ansatz:




                    If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.




                    This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.



                    This argument actually shows




                    $17 mid 29x + 18y iff 17 mid 23x + 9y$




                    Here is an explanation.



                    Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
                    Its determinant is $-153$, which is $0$ mod $17$.
                    Therefore, its two rows are linearly dependent mod $17$.
                    In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)



                    For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 19 '18 at 12:40

























                    answered Dec 19 '18 at 12:18









                    lhflhf

                    166k10171396




                    166k10171396












                    • $begingroup$
                      Adapted from math.stackexchange.com/a/2947271/589
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:18






                    • 1




                      $begingroup$
                      Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:23










                    • $begingroup$
                      @someone, just compute $9(29x + 18y)$ mod $17$.
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:25










                    • $begingroup$
                      Oh, but whats the trick with 9 ?
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:26










                    • $begingroup$
                      @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
                      $endgroup$
                      – Shaun
                      Dec 19 '18 at 12:29


















                    • $begingroup$
                      Adapted from math.stackexchange.com/a/2947271/589
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:18






                    • 1




                      $begingroup$
                      Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:23










                    • $begingroup$
                      @someone, just compute $9(29x + 18y)$ mod $17$.
                      $endgroup$
                      – lhf
                      Dec 19 '18 at 12:25










                    • $begingroup$
                      Oh, but whats the trick with 9 ?
                      $endgroup$
                      – user626177
                      Dec 19 '18 at 12:26










                    • $begingroup$
                      @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
                      $endgroup$
                      – Shaun
                      Dec 19 '18 at 12:29
















                    $begingroup$
                    Adapted from math.stackexchange.com/a/2947271/589
                    $endgroup$
                    – lhf
                    Dec 19 '18 at 12:18




                    $begingroup$
                    Adapted from math.stackexchange.com/a/2947271/589
                    $endgroup$
                    – lhf
                    Dec 19 '18 at 12:18




                    1




                    1




                    $begingroup$
                    Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
                    $endgroup$
                    – user626177
                    Dec 19 '18 at 12:23




                    $begingroup$
                    Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
                    $endgroup$
                    – user626177
                    Dec 19 '18 at 12:23












                    $begingroup$
                    @someone, just compute $9(29x + 18y)$ mod $17$.
                    $endgroup$
                    – lhf
                    Dec 19 '18 at 12:25




                    $begingroup$
                    @someone, just compute $9(29x + 18y)$ mod $17$.
                    $endgroup$
                    – lhf
                    Dec 19 '18 at 12:25












                    $begingroup$
                    Oh, but whats the trick with 9 ?
                    $endgroup$
                    – user626177
                    Dec 19 '18 at 12:26




                    $begingroup$
                    Oh, but whats the trick with 9 ?
                    $endgroup$
                    – user626177
                    Dec 19 '18 at 12:26












                    $begingroup$
                    @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
                    $endgroup$
                    – Shaun
                    Dec 19 '18 at 12:29




                    $begingroup$
                    @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
                    $endgroup$
                    – Shaun
                    Dec 19 '18 at 12:29


















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