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

259514




259514








  • 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.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      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

          855




          855























              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

              1,081724




              1,081724












              • $begingroup$
                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

              3,2521816




              3,2521816












              • $begingroup$
                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

                  3,2521816




                  3,2521816






























                      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

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                      Read more

                      The Binding of Isaac: Rebirth/Afterbirth

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                      Read more

                      Dobbiaco

                      Image
                      Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
                      Read more