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









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      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






























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