Finding integer solution to a quadratic equation in two unknowns












1












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We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.







elementary-number-theory divisibility diophantine-equations







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  • 1




    $begingroup$
    Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
    $endgroup$
    – fleablood
    4 hours ago
















1












$begingroup$



We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.







elementary-number-theory divisibility diophantine-equations







share|cite|improve this question









New contributor




BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
    $endgroup$
    – fleablood
    4 hours ago














1












1








1


1



$begingroup$



We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.







elementary-number-theory divisibility diophantine-equations







share|cite|improve this question









New contributor




BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.







elementary-number-theory divisibility diophantine-equations




elementary-number-theory divisibility diophantine-equations






share|cite|improve this question









New contributor




BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









greedoid

46.3k1160117




46.3k1160117






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asked 4 hours ago









BIDS SalvaterraBIDS Salvaterra

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91




New contributor




BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
    $endgroup$
    – fleablood
    4 hours ago














  • 1




    $begingroup$
    Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
    $endgroup$
    – fleablood
    4 hours ago








1




1




$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
4 hours ago




$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
4 hours ago










3 Answers
3






active

oldest

votes


















6












$begingroup$

Hint $$ (m+n)(m-n)= (m+n)+2018$$



so $$ (m+n)(m-n-1)= 2018$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$



    so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$



    If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$



    Can you finish?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Simpler start: separating variables to either side gives:
      $$m^2-m=n^2+n+2018$$
      which then factors roughly for the variables as:
      $$m(m-1)=n(n+1)+2018$$



      which since both pairs(m,m-1) and (n,n-1) are consecutive integers, you can divide both sides by two giving:



      $$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$



      But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Hint $$ (m+n)(m-n)= (m+n)+2018$$



        so $$ (m+n)(m-n-1)= 2018$$






        share|cite|improve this answer









        $endgroup$


















          6












          $begingroup$

          Hint $$ (m+n)(m-n)= (m+n)+2018$$



          so $$ (m+n)(m-n-1)= 2018$$






          share|cite|improve this answer









          $endgroup$
















            6












            6








            6





            $begingroup$

            Hint $$ (m+n)(m-n)= (m+n)+2018$$



            so $$ (m+n)(m-n-1)= 2018$$






            share|cite|improve this answer









            $endgroup$



            Hint $$ (m+n)(m-n)= (m+n)+2018$$



            so $$ (m+n)(m-n-1)= 2018$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            greedoidgreedoid

            46.3k1160117




            46.3k1160117























                3












                $begingroup$

                Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$



                so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$



                If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$



                Can you finish?






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$



                  so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$



                  If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$



                  Can you finish?






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$



                    so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$



                    If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$



                    Can you finish?






                    share|cite|improve this answer









                    $endgroup$



                    Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$



                    so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$



                    If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$



                    Can you finish?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 4 hours ago









                    greedoidgreedoid

                    46.3k1160117




                    46.3k1160117























                        0












                        $begingroup$

                        Simpler start: separating variables to either side gives:
                        $$m^2-m=n^2+n+2018$$
                        which then factors roughly for the variables as:
                        $$m(m-1)=n(n+1)+2018$$



                        which since both pairs(m,m-1) and (n,n-1) are consecutive integers, you can divide both sides by two giving:



                        $$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$



                        But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Simpler start: separating variables to either side gives:
                          $$m^2-m=n^2+n+2018$$
                          which then factors roughly for the variables as:
                          $$m(m-1)=n(n+1)+2018$$



                          which since both pairs(m,m-1) and (n,n-1) are consecutive integers, you can divide both sides by two giving:



                          $$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$



                          But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Simpler start: separating variables to either side gives:
                            $$m^2-m=n^2+n+2018$$
                            which then factors roughly for the variables as:
                            $$m(m-1)=n(n+1)+2018$$



                            which since both pairs(m,m-1) and (n,n-1) are consecutive integers, you can divide both sides by two giving:



                            $$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$



                            But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .






                            share|cite|improve this answer









                            $endgroup$



                            Simpler start: separating variables to either side gives:
                            $$m^2-m=n^2+n+2018$$
                            which then factors roughly for the variables as:
                            $$m(m-1)=n(n+1)+2018$$



                            which since both pairs(m,m-1) and (n,n-1) are consecutive integers, you can divide both sides by two giving:



                            $$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$



                            But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            Roddy MacPheeRoddy MacPhee

                            22414




                            22414






















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