Find the pairs $(m,k)$ solving the diophantine equation $m^2=7k+9$












0












$begingroup$


Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49
















0












$begingroup$


Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49














0












0








0





$begingroup$


Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers







share|cite|improve this question











$endgroup$




Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers




number-theory diophantine-equations integers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 14:48









Peter

48.5k1139135




48.5k1139135










asked Jan 11 at 12:53









HeartHeart

29518




29518








  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49














  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49








1




1




$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57






$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57






2




2




$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57




$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57












$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49




$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49










3 Answers
3






active

oldest

votes


















2












$begingroup$

$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



For instance, if $m=81=4+77$, then $k=936$, and



$$81^2=7 cdot 936 + 9$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Above equation $m^2=7k+9$ has solution's,



    $m=7w-4$



    $k=(7w-1)(w-1)$, where, 'w' is any integer



    For $w=2$ we get, $(m.k)=(10,13)$



    $(10)^2=7(13)+9$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What about (3,0)?
      $endgroup$
      – pendermath
      Jan 11 at 17:27










    • $begingroup$
      It is included $w=1$
      $endgroup$
      – Mike
      Jan 11 at 18:31










    • $begingroup$
      Sorry, I meant (4,1)
      $endgroup$
      – pendermath
      Jan 11 at 21:04










    • $begingroup$
      For w=0, you will get (m,k)=(4,1)
      $endgroup$
      – Sam
      Jan 11 at 22:10










    • $begingroup$
      If $w=0$, then $m=-4$, not 4, right?
      $endgroup$
      – pendermath
      Jan 12 at 22:37



















    0












    $begingroup$

    As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



    Therefore, you have two sets of solutions:



    Case 1: $7|m-3$ then
    $$m-3=7l \
    m=7l+3 \
    7k+9=m^2=49l^2+42l+9 \
    k=7l^2+6l\
    (k,m)= ( 7l^2+6l, 7l+3)$$



    Case 2: $7|m+3$ then
    $$m+3=7l \
    m=7l-3 \
    7k+9=m^2=49l^2-42l+9 \
    k=7l^2+6l\
    (k,m)= ( 7l^2-6l, 7l-3)$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



      The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



      $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



      For instance, if $m=81=4+77$, then $k=936$, and



      $$81^2=7 cdot 936 + 9$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



        The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



        $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



        For instance, if $m=81=4+77$, then $k=936$, and



        $$81^2=7 cdot 936 + 9$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



          The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



          $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



          For instance, if $m=81=4+77$, then $k=936$, and



          $$81^2=7 cdot 936 + 9$$






          share|cite|improve this answer









          $endgroup$



          $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



          The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



          $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



          For instance, if $m=81=4+77$, then $k=936$, and



          $$81^2=7 cdot 936 + 9$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 17:22









          pendermathpendermath

          58612




          58612























              0












              $begingroup$

              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37
















              0












              $begingroup$

              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37














              0












              0








              0





              $begingroup$

              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$






              share|cite|improve this answer











              $endgroup$



              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 12 at 23:18









              Namaste

              1




              1










              answered Jan 11 at 17:26









              SamSam

              11




              11












              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37


















              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37
















              $begingroup$
              What about (3,0)?
              $endgroup$
              – pendermath
              Jan 11 at 17:27




              $begingroup$
              What about (3,0)?
              $endgroup$
              – pendermath
              Jan 11 at 17:27












              $begingroup$
              It is included $w=1$
              $endgroup$
              – Mike
              Jan 11 at 18:31




              $begingroup$
              It is included $w=1$
              $endgroup$
              – Mike
              Jan 11 at 18:31












              $begingroup$
              Sorry, I meant (4,1)
              $endgroup$
              – pendermath
              Jan 11 at 21:04




              $begingroup$
              Sorry, I meant (4,1)
              $endgroup$
              – pendermath
              Jan 11 at 21:04












              $begingroup$
              For w=0, you will get (m,k)=(4,1)
              $endgroup$
              – Sam
              Jan 11 at 22:10




              $begingroup$
              For w=0, you will get (m,k)=(4,1)
              $endgroup$
              – Sam
              Jan 11 at 22:10












              $begingroup$
              If $w=0$, then $m=-4$, not 4, right?
              $endgroup$
              – pendermath
              Jan 12 at 22:37




              $begingroup$
              If $w=0$, then $m=-4$, not 4, right?
              $endgroup$
              – pendermath
              Jan 12 at 22:37











              0












              $begingroup$

              As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



              Therefore, you have two sets of solutions:



              Case 1: $7|m-3$ then
              $$m-3=7l \
              m=7l+3 \
              7k+9=m^2=49l^2+42l+9 \
              k=7l^2+6l\
              (k,m)= ( 7l^2+6l, 7l+3)$$



              Case 2: $7|m+3$ then
              $$m+3=7l \
              m=7l-3 \
              7k+9=m^2=49l^2-42l+9 \
              k=7l^2+6l\
              (k,m)= ( 7l^2-6l, 7l-3)$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



                Therefore, you have two sets of solutions:



                Case 1: $7|m-3$ then
                $$m-3=7l \
                m=7l+3 \
                7k+9=m^2=49l^2+42l+9 \
                k=7l^2+6l\
                (k,m)= ( 7l^2+6l, 7l+3)$$



                Case 2: $7|m+3$ then
                $$m+3=7l \
                m=7l-3 \
                7k+9=m^2=49l^2-42l+9 \
                k=7l^2+6l\
                (k,m)= ( 7l^2-6l, 7l-3)$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



                  Therefore, you have two sets of solutions:



                  Case 1: $7|m-3$ then
                  $$m-3=7l \
                  m=7l+3 \
                  7k+9=m^2=49l^2+42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2+6l, 7l+3)$$



                  Case 2: $7|m+3$ then
                  $$m+3=7l \
                  m=7l-3 \
                  7k+9=m^2=49l^2-42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2-6l, 7l-3)$$






                  share|cite|improve this answer









                  $endgroup$



                  As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



                  Therefore, you have two sets of solutions:



                  Case 1: $7|m-3$ then
                  $$m-3=7l \
                  m=7l+3 \
                  7k+9=m^2=49l^2+42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2+6l, 7l+3)$$



                  Case 2: $7|m+3$ then
                  $$m+3=7l \
                  m=7l-3 \
                  7k+9=m^2=49l^2-42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2-6l, 7l-3)$$







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                  answered Jan 12 at 23:45









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