How to solve these two simultaneous “divisibilities” : $n+1mid m^2+1$ and $m+1mid n^2+1$












83












$begingroup$


Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?



I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.



Thanks !






number-theory elementary-number-theory divisibility diophantine-equations congruences







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    Could you write down your proof for the infinite number of solutions? It may help everyone.
    $endgroup$
    – user37238
    Feb 25 '16 at 8:22






  • 3




    $begingroup$
    If you understand French, there is a long discussion (partly numerical, partly theoretical) here
    $endgroup$
    – Giovanni Resta
    Feb 25 '16 at 8:28






  • 2




    $begingroup$
    Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
    $endgroup$
    – uvdose
    Feb 25 '16 at 8:37








  • 4




    $begingroup$
    @individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
    $endgroup$
    – uvdose
    Feb 29 '16 at 14:25








  • 5




    $begingroup$
    If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
    $endgroup$
    – 201044
    Mar 15 '16 at 23:39
















83












$begingroup$


Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?



I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.



Thanks !






number-theory elementary-number-theory divisibility diophantine-equations congruences







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    Could you write down your proof for the infinite number of solutions? It may help everyone.
    $endgroup$
    – user37238
    Feb 25 '16 at 8:22






  • 3




    $begingroup$
    If you understand French, there is a long discussion (partly numerical, partly theoretical) here
    $endgroup$
    – Giovanni Resta
    Feb 25 '16 at 8:28






  • 2




    $begingroup$
    Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
    $endgroup$
    – uvdose
    Feb 25 '16 at 8:37








  • 4




    $begingroup$
    @individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
    $endgroup$
    – uvdose
    Feb 29 '16 at 14:25








  • 5




    $begingroup$
    If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
    $endgroup$
    – 201044
    Mar 15 '16 at 23:39














83












83








83


41



$begingroup$


Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?



I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.



Thanks !






number-theory elementary-number-theory divisibility diophantine-equations congruences







share|cite|improve this question











$endgroup$




Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?



I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.



Thanks !






number-theory elementary-number-theory divisibility diophantine-equations congruences




number-theory elementary-number-theory divisibility diophantine-equations congruences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '16 at 16:56







uvdose

















asked Feb 25 '16 at 7:49









uvdoseuvdose

299615




299615








  • 12




    $begingroup$
    Could you write down your proof for the infinite number of solutions? It may help everyone.
    $endgroup$
    – user37238
    Feb 25 '16 at 8:22






  • 3




    $begingroup$
    If you understand French, there is a long discussion (partly numerical, partly theoretical) here
    $endgroup$
    – Giovanni Resta
    Feb 25 '16 at 8:28






  • 2




    $begingroup$
    Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
    $endgroup$
    – uvdose
    Feb 25 '16 at 8:37








  • 4




    $begingroup$
    @individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
    $endgroup$
    – uvdose
    Feb 29 '16 at 14:25








  • 5




    $begingroup$
    If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
    $endgroup$
    – 201044
    Mar 15 '16 at 23:39














  • 12




    $begingroup$
    Could you write down your proof for the infinite number of solutions? It may help everyone.
    $endgroup$
    – user37238
    Feb 25 '16 at 8:22






  • 3




    $begingroup$
    If you understand French, there is a long discussion (partly numerical, partly theoretical) here
    $endgroup$
    – Giovanni Resta
    Feb 25 '16 at 8:28






  • 2




    $begingroup$
    Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
    $endgroup$
    – uvdose
    Feb 25 '16 at 8:37








  • 4




    $begingroup$
    @individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
    $endgroup$
    – uvdose
    Feb 29 '16 at 14:25








  • 5




    $begingroup$
    If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
    $endgroup$
    – 201044
    Mar 15 '16 at 23:39








12




12




$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22




$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22




3




3




$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28




$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28




2




2




$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37






$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37






4




4




$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25






$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25






5




5




$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39




$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39










4 Answers
4






active

oldest

votes


















3












$begingroup$

Some further results along the lines of thought of @individ:



Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-dcdot p^2+s^2=1$$
Then,
begin{align}
m &= acdot p^2+bcdot pq +ccdot q^2\
n &= acdot p^2-bcdot pq +ccdot q^2
end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
begin{align}
(10,4,-2,-15)\
(39,12,-3,-65)\
end{align}
Sadly, the solutions are negative.



Here are some examples:
begin{align}
(m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
(m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
(m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
(m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
end{align}
P.S. I am also very curious how @individ thought of this parametrization.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Trying again, now as answer instead of just a comment:



    I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?



    The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.



    simultaneouos_divisions_MSE.pdf






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      why link pdf? why not just list answer here?
      $endgroup$
      – qwr
      Jan 9 at 20:34



















    0












    $begingroup$

    You can record a similar system:



    $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



    Parametrization of solutions we write this.



    $$m=q(3x-q)$$



    $$n=2x^2-qx-q^2$$



    $$t=3q^2-3xq+2x^2$$



    $$z=5q^2-2qx+x^2$$



    $$k=5q^2-8qx+4x^2$$



    Consider a special case.



    $$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$



    Using the solutions of the equation Pell.



    $$p^2-15s^2=1$$



    Enough to know first, everything else will find a formula. $(p;s) - (4;1)$



    $$p_2=4p+15s$$



    $$s_2=p+4s$$



    The solution then write.



    $$m=-2p^2-4ps+10s^2$$



    $$n=-2p^2+4ps+10s^2$$



    $$z=8m+9-n$$



    $$k=8n+9-m$$



    These solutions are negative.



    And a positive decision of the same are determined by the Pell equation.



    $$p^2-65s^2=-1$$



    Use the first solution. $(p;s) - (8;1)$



    Next find the formula.



    $$p_2=129p+1040s$$



    $$s_2=16p+129s$$



    Will make a replacement.



    $$x=p^2+6ps+13s^2$$



    $$y=p^2-6ps+13s^2$$



    The decision record.



    $$m=2x-1$$



    $$n=2y-1$$



    $$z=9x-2y+2$$



    $$k=9y-2x+2$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
      $endgroup$
      – MikeTeX
      Jan 14 at 15:50












    • $begingroup$
      @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
      $endgroup$
      – individ
      Jan 14 at 16:18





















    -1












    $begingroup$

    To solve this system of equations - it is necessary to solve the system.



    $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



    It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.



    $$m=33643p^2pm5404ps+217s^2$$



    $$n=5491p^2pm852ps+33s^2$$



    $$t=s^2-153p^2$$



    $$z=212041p^2pm34274ps+1385s^2$$



    $$k=901p^2pm134ps+5s^2$$



    We need a case of when. $t=s^2-153p^2=1$



    Knowing the first decision $(s ; p ) - (2177;176)$



    The rest can be found by the formula.



    $$s=2177s+26928p$$



    $$p=176s+2177p$$



    Although this equation can be not enough. We need to find when there are multiple solutions.






    share|cite|improve this answer









    $endgroup$














      Your Answer








      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1671427%2fhow-to-solve-these-two-simultaneous-divisibilities-n1-mid-m21-and-m1%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Some further results along the lines of thought of @individ:



      Suppose $p$ and $s$ are solutions to the Pell's equation:
      $$-dcdot p^2+s^2=1$$
      Then,
      begin{align}
      m &= acdot p^2+bcdot pq +ccdot q^2\
      n &= acdot p^2-bcdot pq +ccdot q^2
      end{align}
      are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
      begin{align}
      (10,4,-2,-15)\
      (39,12,-3,-65)\
      end{align}
      Sadly, the solutions are negative.



      Here are some examples:
      begin{align}
      (m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
      (m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
      (m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
      (m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
      (m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
      end{align}
      P.S. I am also very curious how @individ thought of this parametrization.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Some further results along the lines of thought of @individ:



        Suppose $p$ and $s$ are solutions to the Pell's equation:
        $$-dcdot p^2+s^2=1$$
        Then,
        begin{align}
        m &= acdot p^2+bcdot pq +ccdot q^2\
        n &= acdot p^2-bcdot pq +ccdot q^2
        end{align}
        are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
        begin{align}
        (10,4,-2,-15)\
        (39,12,-3,-65)\
        end{align}
        Sadly, the solutions are negative.



        Here are some examples:
        begin{align}
        (m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
        (m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
        (m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
        (m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
        (m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
        end{align}
        P.S. I am also very curious how @individ thought of this parametrization.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Some further results along the lines of thought of @individ:



          Suppose $p$ and $s$ are solutions to the Pell's equation:
          $$-dcdot p^2+s^2=1$$
          Then,
          begin{align}
          m &= acdot p^2+bcdot pq +ccdot q^2\
          n &= acdot p^2-bcdot pq +ccdot q^2
          end{align}
          are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
          begin{align}
          (10,4,-2,-15)\
          (39,12,-3,-65)\
          end{align}
          Sadly, the solutions are negative.



          Here are some examples:
          begin{align}
          (m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
          (m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
          (m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
          (m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
          (m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
          end{align}
          P.S. I am also very curious how @individ thought of this parametrization.






          share|cite|improve this answer









          $endgroup$



          Some further results along the lines of thought of @individ:



          Suppose $p$ and $s$ are solutions to the Pell's equation:
          $$-dcdot p^2+s^2=1$$
          Then,
          begin{align}
          m &= acdot p^2+bcdot pq +ccdot q^2\
          n &= acdot p^2-bcdot pq +ccdot q^2
          end{align}
          are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
          begin{align}
          (10,4,-2,-15)\
          (39,12,-3,-65)\
          end{align}
          Sadly, the solutions are negative.



          Here are some examples:
          begin{align}
          (m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
          (m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
          (m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
          (m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
          (m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
          end{align}
          P.S. I am also very curious how @individ thought of this parametrization.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '16 at 13:58









          Yifan ZhuYifan Zhu

          905




          905























              1












              $begingroup$

              Trying again, now as answer instead of just a comment:



              I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?



              The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.



              simultaneouos_divisions_MSE.pdf






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                why link pdf? why not just list answer here?
                $endgroup$
                – qwr
                Jan 9 at 20:34
















              1












              $begingroup$

              Trying again, now as answer instead of just a comment:



              I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?



              The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.



              simultaneouos_divisions_MSE.pdf






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                why link pdf? why not just list answer here?
                $endgroup$
                – qwr
                Jan 9 at 20:34














              1












              1








              1





              $begingroup$

              Trying again, now as answer instead of just a comment:



              I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?



              The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.



              simultaneouos_divisions_MSE.pdf






              share|cite|improve this answer











              $endgroup$



              Trying again, now as answer instead of just a comment:



              I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?



              The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.



              simultaneouos_divisions_MSE.pdf







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 22 '18 at 21:58

























              answered Dec 22 '18 at 15:47









              Maestro13Maestro13

              1,106724




              1,106724












              • $begingroup$
                why link pdf? why not just list answer here?
                $endgroup$
                – qwr
                Jan 9 at 20:34


















              • $begingroup$
                why link pdf? why not just list answer here?
                $endgroup$
                – qwr
                Jan 9 at 20:34
















              $begingroup$
              why link pdf? why not just list answer here?
              $endgroup$
              – qwr
              Jan 9 at 20:34




              $begingroup$
              why link pdf? why not just list answer here?
              $endgroup$
              – qwr
              Jan 9 at 20:34











              0












              $begingroup$

              You can record a similar system:



              $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



              Parametrization of solutions we write this.



              $$m=q(3x-q)$$



              $$n=2x^2-qx-q^2$$



              $$t=3q^2-3xq+2x^2$$



              $$z=5q^2-2qx+x^2$$



              $$k=5q^2-8qx+4x^2$$



              Consider a special case.



              $$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$



              Using the solutions of the equation Pell.



              $$p^2-15s^2=1$$



              Enough to know first, everything else will find a formula. $(p;s) - (4;1)$



              $$p_2=4p+15s$$



              $$s_2=p+4s$$



              The solution then write.



              $$m=-2p^2-4ps+10s^2$$



              $$n=-2p^2+4ps+10s^2$$



              $$z=8m+9-n$$



              $$k=8n+9-m$$



              These solutions are negative.



              And a positive decision of the same are determined by the Pell equation.



              $$p^2-65s^2=-1$$



              Use the first solution. $(p;s) - (8;1)$



              Next find the formula.



              $$p_2=129p+1040s$$



              $$s_2=16p+129s$$



              Will make a replacement.



              $$x=p^2+6ps+13s^2$$



              $$y=p^2-6ps+13s^2$$



              The decision record.



              $$m=2x-1$$



              $$n=2y-1$$



              $$z=9x-2y+2$$



              $$k=9y-2x+2$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
                $endgroup$
                – MikeTeX
                Jan 14 at 15:50












              • $begingroup$
                @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
                $endgroup$
                – individ
                Jan 14 at 16:18


















              0












              $begingroup$

              You can record a similar system:



              $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



              Parametrization of solutions we write this.



              $$m=q(3x-q)$$



              $$n=2x^2-qx-q^2$$



              $$t=3q^2-3xq+2x^2$$



              $$z=5q^2-2qx+x^2$$



              $$k=5q^2-8qx+4x^2$$



              Consider a special case.



              $$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$



              Using the solutions of the equation Pell.



              $$p^2-15s^2=1$$



              Enough to know first, everything else will find a formula. $(p;s) - (4;1)$



              $$p_2=4p+15s$$



              $$s_2=p+4s$$



              The solution then write.



              $$m=-2p^2-4ps+10s^2$$



              $$n=-2p^2+4ps+10s^2$$



              $$z=8m+9-n$$



              $$k=8n+9-m$$



              These solutions are negative.



              And a positive decision of the same are determined by the Pell equation.



              $$p^2-65s^2=-1$$



              Use the first solution. $(p;s) - (8;1)$



              Next find the formula.



              $$p_2=129p+1040s$$



              $$s_2=16p+129s$$



              Will make a replacement.



              $$x=p^2+6ps+13s^2$$



              $$y=p^2-6ps+13s^2$$



              The decision record.



              $$m=2x-1$$



              $$n=2y-1$$



              $$z=9x-2y+2$$



              $$k=9y-2x+2$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
                $endgroup$
                – MikeTeX
                Jan 14 at 15:50












              • $begingroup$
                @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
                $endgroup$
                – individ
                Jan 14 at 16:18
















              0












              0








              0





              $begingroup$

              You can record a similar system:



              $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



              Parametrization of solutions we write this.



              $$m=q(3x-q)$$



              $$n=2x^2-qx-q^2$$



              $$t=3q^2-3xq+2x^2$$



              $$z=5q^2-2qx+x^2$$



              $$k=5q^2-8qx+4x^2$$



              Consider a special case.



              $$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$



              Using the solutions of the equation Pell.



              $$p^2-15s^2=1$$



              Enough to know first, everything else will find a formula. $(p;s) - (4;1)$



              $$p_2=4p+15s$$



              $$s_2=p+4s$$



              The solution then write.



              $$m=-2p^2-4ps+10s^2$$



              $$n=-2p^2+4ps+10s^2$$



              $$z=8m+9-n$$



              $$k=8n+9-m$$



              These solutions are negative.



              And a positive decision of the same are determined by the Pell equation.



              $$p^2-65s^2=-1$$



              Use the first solution. $(p;s) - (8;1)$



              Next find the formula.



              $$p_2=129p+1040s$$



              $$s_2=16p+129s$$



              Will make a replacement.



              $$x=p^2+6ps+13s^2$$



              $$y=p^2-6ps+13s^2$$



              The decision record.



              $$m=2x-1$$



              $$n=2y-1$$



              $$z=9x-2y+2$$



              $$k=9y-2x+2$$






              share|cite|improve this answer











              $endgroup$



              You can record a similar system:



              $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



              Parametrization of solutions we write this.



              $$m=q(3x-q)$$



              $$n=2x^2-qx-q^2$$



              $$t=3q^2-3xq+2x^2$$



              $$z=5q^2-2qx+x^2$$



              $$k=5q^2-8qx+4x^2$$



              Consider a special case.



              $$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$



              Using the solutions of the equation Pell.



              $$p^2-15s^2=1$$



              Enough to know first, everything else will find a formula. $(p;s) - (4;1)$



              $$p_2=4p+15s$$



              $$s_2=p+4s$$



              The solution then write.



              $$m=-2p^2-4ps+10s^2$$



              $$n=-2p^2+4ps+10s^2$$



              $$z=8m+9-n$$



              $$k=8n+9-m$$



              These solutions are negative.



              And a positive decision of the same are determined by the Pell equation.



              $$p^2-65s^2=-1$$



              Use the first solution. $(p;s) - (8;1)$



              Next find the formula.



              $$p_2=129p+1040s$$



              $$s_2=16p+129s$$



              Will make a replacement.



              $$x=p^2+6ps+13s^2$$



              $$y=p^2-6ps+13s^2$$



              The decision record.



              $$m=2x-1$$



              $$n=2y-1$$



              $$z=9x-2y+2$$



              $$k=9y-2x+2$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 14 '17 at 12:45

























              answered Dec 30 '16 at 8:44









              individindivid

              3,2621916




              3,2621916












              • $begingroup$
                In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
                $endgroup$
                – MikeTeX
                Jan 14 at 15:50












              • $begingroup$
                @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
                $endgroup$
                – individ
                Jan 14 at 16:18




















              • $begingroup$
                In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
                $endgroup$
                – MikeTeX
                Jan 14 at 15:50












              • $begingroup$
                @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
                $endgroup$
                – individ
                Jan 14 at 16:18


















              $begingroup$
              In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
              $endgroup$
              – MikeTeX
              Jan 14 at 15:50






              $begingroup$
              In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
              $endgroup$
              – MikeTeX
              Jan 14 at 15:50














              $begingroup$
              @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
              $endgroup$
              – individ
              Jan 14 at 16:18






              $begingroup$
              @MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
              $endgroup$
              – individ
              Jan 14 at 16:18













              -1












              $begingroup$

              To solve this system of equations - it is necessary to solve the system.



              $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



              It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.



              $$m=33643p^2pm5404ps+217s^2$$



              $$n=5491p^2pm852ps+33s^2$$



              $$t=s^2-153p^2$$



              $$z=212041p^2pm34274ps+1385s^2$$



              $$k=901p^2pm134ps+5s^2$$



              We need a case of when. $t=s^2-153p^2=1$



              Knowing the first decision $(s ; p ) - (2177;176)$



              The rest can be found by the formula.



              $$s=2177s+26928p$$



              $$p=176s+2177p$$



              Although this equation can be not enough. We need to find when there are multiple solutions.






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                To solve this system of equations - it is necessary to solve the system.



                $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



                It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.



                $$m=33643p^2pm5404ps+217s^2$$



                $$n=5491p^2pm852ps+33s^2$$



                $$t=s^2-153p^2$$



                $$z=212041p^2pm34274ps+1385s^2$$



                $$k=901p^2pm134ps+5s^2$$



                We need a case of when. $t=s^2-153p^2=1$



                Knowing the first decision $(s ; p ) - (2177;176)$



                The rest can be found by the formula.



                $$s=2177s+26928p$$



                $$p=176s+2177p$$



                Although this equation can be not enough. We need to find when there are multiple solutions.






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  To solve this system of equations - it is necessary to solve the system.



                  $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



                  It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.



                  $$m=33643p^2pm5404ps+217s^2$$



                  $$n=5491p^2pm852ps+33s^2$$



                  $$t=s^2-153p^2$$



                  $$z=212041p^2pm34274ps+1385s^2$$



                  $$k=901p^2pm134ps+5s^2$$



                  We need a case of when. $t=s^2-153p^2=1$



                  Knowing the first decision $(s ; p ) - (2177;176)$



                  The rest can be found by the formula.



                  $$s=2177s+26928p$$



                  $$p=176s+2177p$$



                  Although this equation can be not enough. We need to find when there are multiple solutions.






                  share|cite|improve this answer









                  $endgroup$



                  To solve this system of equations - it is necessary to solve the system.



                  $$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$



                  It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.



                  $$m=33643p^2pm5404ps+217s^2$$



                  $$n=5491p^2pm852ps+33s^2$$



                  $$t=s^2-153p^2$$



                  $$z=212041p^2pm34274ps+1385s^2$$



                  $$k=901p^2pm134ps+5s^2$$



                  We need a case of when. $t=s^2-153p^2=1$



                  Knowing the first decision $(s ; p ) - (2177;176)$



                  The rest can be found by the formula.



                  $$s=2177s+26928p$$



                  $$p=176s+2177p$$



                  Although this equation can be not enough. We need to find when there are multiple solutions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 '17 at 16:57









                  individindivid

                  3,2621916




                  3,2621916






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1671427%2fhow-to-solve-these-two-simultaneous-divisibilities-n1-mid-m21-and-m1%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
                      Read more