Find all integer solutions of $3(m^2 + n^2) - 7(m+n) = -4$.












1














The solution and further information about how to solve this type of equation about how to solve this type of lattice point and circles will be much appreciated. Thanks in advance.






circle diophantine-equations







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  • 2




    Multiply both sides by $12$ and then complete squares.
    – saulspatz
    Jan 4 at 17:30
















1














The solution and further information about how to solve this type of equation about how to solve this type of lattice point and circles will be much appreciated. Thanks in advance.






circle diophantine-equations







share|cite|improve this question


















  • 2




    Multiply both sides by $12$ and then complete squares.
    – saulspatz
    Jan 4 at 17:30














1












1








1







The solution and further information about how to solve this type of equation about how to solve this type of lattice point and circles will be much appreciated. Thanks in advance.






circle diophantine-equations







share|cite|improve this question













The solution and further information about how to solve this type of equation about how to solve this type of lattice point and circles will be much appreciated. Thanks in advance.






circle diophantine-equations




circle diophantine-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 4 at 17:22









Shafin AhmedShafin Ahmed

345




345








  • 2




    Multiply both sides by $12$ and then complete squares.
    – saulspatz
    Jan 4 at 17:30














  • 2




    Multiply both sides by $12$ and then complete squares.
    – saulspatz
    Jan 4 at 17:30








2




2




Multiply both sides by $12$ and then complete squares.
– saulspatz
Jan 4 at 17:30




Multiply both sides by $12$ and then complete squares.
– saulspatz
Jan 4 at 17:30










3 Answers
3






active

oldest

votes


















1














First, expand the equation and bring the $-4$ to the other side.



$$3m^2+3n^2-7m-7n+4 = 0$$



To solve for $m$, you can rearrange the LHS:



$$3m^2-7m+left(3n^2-7n+4right) = 0$$



By the Quadratic Formula,



$$m = frac{-(-7)pmsqrt{(-7)^2-4(3)left(3n^2-7n+4right)}}{2(3)}$$



$$m = frac{7pmsqrt{1-36n^2+84n}}{6}$$



You want the discriminant to be a perfect square:



$$1-36n^2+84n = t^2$$



and



$$1-36n^2+84n geq 0$$






share|cite|improve this answer





























    1














    Hint:



    $$3m^2-7m+3n^2-7n+4=0$$



    What is the discriminant of the quadratic equation in $m$






    share|cite|improve this answer





























      1














      Following the comment by @saulspatz, we get
      $$
      (6m-7)^2+(6n-7)^2=50
      $$
      So, we have to solve $a^2+b^2=50$ with $a equiv b equiv -7 bmod 6$.



      The solutions are
      $$
      begin{array}{cccc}
      a & b & m & n \
      -7 & -1 & 0 & 1 \
      -1 & -7 & 1 & 0 \
      5 & 5 & 2 & 2 \
      end{array}
      $$
      These were found by brute force. For numbers larger than $50$, consider its prime factorization and use the Brahmagupta–Fibonacci identity.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        First, expand the equation and bring the $-4$ to the other side.



        $$3m^2+3n^2-7m-7n+4 = 0$$



        To solve for $m$, you can rearrange the LHS:



        $$3m^2-7m+left(3n^2-7n+4right) = 0$$



        By the Quadratic Formula,



        $$m = frac{-(-7)pmsqrt{(-7)^2-4(3)left(3n^2-7n+4right)}}{2(3)}$$



        $$m = frac{7pmsqrt{1-36n^2+84n}}{6}$$



        You want the discriminant to be a perfect square:



        $$1-36n^2+84n = t^2$$



        and



        $$1-36n^2+84n geq 0$$






        share|cite|improve this answer


























          1














          First, expand the equation and bring the $-4$ to the other side.



          $$3m^2+3n^2-7m-7n+4 = 0$$



          To solve for $m$, you can rearrange the LHS:



          $$3m^2-7m+left(3n^2-7n+4right) = 0$$



          By the Quadratic Formula,



          $$m = frac{-(-7)pmsqrt{(-7)^2-4(3)left(3n^2-7n+4right)}}{2(3)}$$



          $$m = frac{7pmsqrt{1-36n^2+84n}}{6}$$



          You want the discriminant to be a perfect square:



          $$1-36n^2+84n = t^2$$



          and



          $$1-36n^2+84n geq 0$$






          share|cite|improve this answer
























            1












            1








            1






            First, expand the equation and bring the $-4$ to the other side.



            $$3m^2+3n^2-7m-7n+4 = 0$$



            To solve for $m$, you can rearrange the LHS:



            $$3m^2-7m+left(3n^2-7n+4right) = 0$$



            By the Quadratic Formula,



            $$m = frac{-(-7)pmsqrt{(-7)^2-4(3)left(3n^2-7n+4right)}}{2(3)}$$



            $$m = frac{7pmsqrt{1-36n^2+84n}}{6}$$



            You want the discriminant to be a perfect square:



            $$1-36n^2+84n = t^2$$



            and



            $$1-36n^2+84n geq 0$$






            share|cite|improve this answer












            First, expand the equation and bring the $-4$ to the other side.



            $$3m^2+3n^2-7m-7n+4 = 0$$



            To solve for $m$, you can rearrange the LHS:



            $$3m^2-7m+left(3n^2-7n+4right) = 0$$



            By the Quadratic Formula,



            $$m = frac{-(-7)pmsqrt{(-7)^2-4(3)left(3n^2-7n+4right)}}{2(3)}$$



            $$m = frac{7pmsqrt{1-36n^2+84n}}{6}$$



            You want the discriminant to be a perfect square:



            $$1-36n^2+84n = t^2$$



            and



            $$1-36n^2+84n geq 0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 17:42









            KM101KM101

            5,7411423




            5,7411423























                1














                Hint:



                $$3m^2-7m+3n^2-7n+4=0$$



                What is the discriminant of the quadratic equation in $m$






                share|cite|improve this answer


























                  1














                  Hint:



                  $$3m^2-7m+3n^2-7n+4=0$$



                  What is the discriminant of the quadratic equation in $m$






                  share|cite|improve this answer
























                    1












                    1








                    1






                    Hint:



                    $$3m^2-7m+3n^2-7n+4=0$$



                    What is the discriminant of the quadratic equation in $m$






                    share|cite|improve this answer












                    Hint:



                    $$3m^2-7m+3n^2-7n+4=0$$



                    What is the discriminant of the quadratic equation in $m$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 17:30









                    lab bhattacharjeelab bhattacharjee

                    224k15156274




                    224k15156274























                        1














                        Following the comment by @saulspatz, we get
                        $$
                        (6m-7)^2+(6n-7)^2=50
                        $$
                        So, we have to solve $a^2+b^2=50$ with $a equiv b equiv -7 bmod 6$.



                        The solutions are
                        $$
                        begin{array}{cccc}
                        a & b & m & n \
                        -7 & -1 & 0 & 1 \
                        -1 & -7 & 1 & 0 \
                        5 & 5 & 2 & 2 \
                        end{array}
                        $$
                        These were found by brute force. For numbers larger than $50$, consider its prime factorization and use the Brahmagupta–Fibonacci identity.






                        share|cite|improve this answer


























                          1














                          Following the comment by @saulspatz, we get
                          $$
                          (6m-7)^2+(6n-7)^2=50
                          $$
                          So, we have to solve $a^2+b^2=50$ with $a equiv b equiv -7 bmod 6$.



                          The solutions are
                          $$
                          begin{array}{cccc}
                          a & b & m & n \
                          -7 & -1 & 0 & 1 \
                          -1 & -7 & 1 & 0 \
                          5 & 5 & 2 & 2 \
                          end{array}
                          $$
                          These were found by brute force. For numbers larger than $50$, consider its prime factorization and use the Brahmagupta–Fibonacci identity.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Following the comment by @saulspatz, we get
                            $$
                            (6m-7)^2+(6n-7)^2=50
                            $$
                            So, we have to solve $a^2+b^2=50$ with $a equiv b equiv -7 bmod 6$.



                            The solutions are
                            $$
                            begin{array}{cccc}
                            a & b & m & n \
                            -7 & -1 & 0 & 1 \
                            -1 & -7 & 1 & 0 \
                            5 & 5 & 2 & 2 \
                            end{array}
                            $$
                            These were found by brute force. For numbers larger than $50$, consider its prime factorization and use the Brahmagupta–Fibonacci identity.






                            share|cite|improve this answer












                            Following the comment by @saulspatz, we get
                            $$
                            (6m-7)^2+(6n-7)^2=50
                            $$
                            So, we have to solve $a^2+b^2=50$ with $a equiv b equiv -7 bmod 6$.



                            The solutions are
                            $$
                            begin{array}{cccc}
                            a & b & m & n \
                            -7 & -1 & 0 & 1 \
                            -1 & -7 & 1 & 0 \
                            5 & 5 & 2 & 2 \
                            end{array}
                            $$
                            These were found by brute force. For numbers larger than $50$, consider its prime factorization and use the Brahmagupta–Fibonacci identity.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 4 at 18:02









                            lhflhf

                            163k10167388




                            163k10167388






























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