Find all integer solutions for $2(x^2+y^2)+x+y=5xy$












4












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Find all integer solutions for $2(x^2+y^2)+x+y=5xy$




I have been attempting to solve this question for a long period of time but have never achieved anything. I tried to go back to WolframAlpha and it gave me that the integer solutions were $x=y=2, x=y=0$. I tried to make to factor it and have a number left on the RHS so I can find out its factors but could not factor it. I also tried making it into the form of: $(x-a_1)^2+(y-a_2)^2+(text{ })(text{ })$ but was unable to determine what would be located inside the empty brackets and the values of $a_1,a_2$. I also tried to multiply the equation by $2$ to get $4x^2=(2x)^2$ Another attempt was assuming that WLOG $xge y iff x=y+a$ which would give me that $a=0$ and from there I could get that $x=y$ and therefore $x=y=0,2$ from basic math. I hope I could get help on this question and thank you anyways.






algebra-precalculus







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    $begingroup$



    Find all integer solutions for $2(x^2+y^2)+x+y=5xy$




    I have been attempting to solve this question for a long period of time but have never achieved anything. I tried to go back to WolframAlpha and it gave me that the integer solutions were $x=y=2, x=y=0$. I tried to make to factor it and have a number left on the RHS so I can find out its factors but could not factor it. I also tried making it into the form of: $(x-a_1)^2+(y-a_2)^2+(text{ })(text{ })$ but was unable to determine what would be located inside the empty brackets and the values of $a_1,a_2$. I also tried to multiply the equation by $2$ to get $4x^2=(2x)^2$ Another attempt was assuming that WLOG $xge y iff x=y+a$ which would give me that $a=0$ and from there I could get that $x=y$ and therefore $x=y=0,2$ from basic math. I hope I could get help on this question and thank you anyways.






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$



      Find all integer solutions for $2(x^2+y^2)+x+y=5xy$




      I have been attempting to solve this question for a long period of time but have never achieved anything. I tried to go back to WolframAlpha and it gave me that the integer solutions were $x=y=2, x=y=0$. I tried to make to factor it and have a number left on the RHS so I can find out its factors but could not factor it. I also tried making it into the form of: $(x-a_1)^2+(y-a_2)^2+(text{ })(text{ })$ but was unable to determine what would be located inside the empty brackets and the values of $a_1,a_2$. I also tried to multiply the equation by $2$ to get $4x^2=(2x)^2$ Another attempt was assuming that WLOG $xge y iff x=y+a$ which would give me that $a=0$ and from there I could get that $x=y$ and therefore $x=y=0,2$ from basic math. I hope I could get help on this question and thank you anyways.






      algebra-precalculus







      share|cite|improve this question









      $endgroup$





      Find all integer solutions for $2(x^2+y^2)+x+y=5xy$




      I have been attempting to solve this question for a long period of time but have never achieved anything. I tried to go back to WolframAlpha and it gave me that the integer solutions were $x=y=2, x=y=0$. I tried to make to factor it and have a number left on the RHS so I can find out its factors but could not factor it. I also tried making it into the form of: $(x-a_1)^2+(y-a_2)^2+(text{ })(text{ })$ but was unable to determine what would be located inside the empty brackets and the values of $a_1,a_2$. I also tried to multiply the equation by $2$ to get $4x^2=(2x)^2$ Another attempt was assuming that WLOG $xge y iff x=y+a$ which would give me that $a=0$ and from there I could get that $x=y$ and therefore $x=y=0,2$ from basic math. I hope I could get help on this question and thank you anyways.






      algebra-precalculus




      algebra-precalculus






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      asked Dec 31 '18 at 18:11









      user587054user587054

      51911




      51911






















          3 Answers
          3






          active

          oldest

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          2












          $begingroup$

          Hint



          $$2x^2+x(1-5y)+2y^2+y=0$$ is a quadratic equation in $x$



          The discriminant $d=(1-5y)^2-8(2y^2+y)=9y^2-18y+1=(3y-3)^2-8$



          As $d$ needs to be perfect square, $(3y-3)^2-8=z^2$(say)



          $(3y-3-z)(3y-3+z)=8$



          As the multiplicands have the same parity, both must be even



          $dfrac{3y-3-z}2cdotdfrac{3y-3+z}2=2$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $$2=(-2)(-1)=2cdot1$$
            $endgroup$
            – lab bhattacharjee
            Dec 31 '18 at 18:21



















          4












          $begingroup$

          $$ (2x-y-1)(x-2y+1) = -1 $$
          With integers, you must have either $1 cdot (-1)$ or $(-1) cdot 1$
          Both cases give an intersection of a pair of (non-parallel) lines, so just two points total






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You have a quadratic equation on $x$ with parameter $y$: $$2x^2+x(1-5y)+2y^2+y=0$$
            so the discriminat must be prefect square
            $$ d^2 = (1-5y)^2-8(2y^2+y) = 9y^2-18y+1$$



            So we have $$8 = 9(y-1)^2 -d^2 = (3y-3-d)(3y-3+d)$$



            and now should be easy...






            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









              2












              $begingroup$

              Hint



              $$2x^2+x(1-5y)+2y^2+y=0$$ is a quadratic equation in $x$



              The discriminant $d=(1-5y)^2-8(2y^2+y)=9y^2-18y+1=(3y-3)^2-8$



              As $d$ needs to be perfect square, $(3y-3)^2-8=z^2$(say)



              $(3y-3-z)(3y-3+z)=8$



              As the multiplicands have the same parity, both must be even



              $dfrac{3y-3-z}2cdotdfrac{3y-3+z}2=2$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                $$2=(-2)(-1)=2cdot1$$
                $endgroup$
                – lab bhattacharjee
                Dec 31 '18 at 18:21
















              2












              $begingroup$

              Hint



              $$2x^2+x(1-5y)+2y^2+y=0$$ is a quadratic equation in $x$



              The discriminant $d=(1-5y)^2-8(2y^2+y)=9y^2-18y+1=(3y-3)^2-8$



              As $d$ needs to be perfect square, $(3y-3)^2-8=z^2$(say)



              $(3y-3-z)(3y-3+z)=8$



              As the multiplicands have the same parity, both must be even



              $dfrac{3y-3-z}2cdotdfrac{3y-3+z}2=2$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                $$2=(-2)(-1)=2cdot1$$
                $endgroup$
                – lab bhattacharjee
                Dec 31 '18 at 18:21














              2












              2








              2





              $begingroup$

              Hint



              $$2x^2+x(1-5y)+2y^2+y=0$$ is a quadratic equation in $x$



              The discriminant $d=(1-5y)^2-8(2y^2+y)=9y^2-18y+1=(3y-3)^2-8$



              As $d$ needs to be perfect square, $(3y-3)^2-8=z^2$(say)



              $(3y-3-z)(3y-3+z)=8$



              As the multiplicands have the same parity, both must be even



              $dfrac{3y-3-z}2cdotdfrac{3y-3+z}2=2$






              share|cite|improve this answer









              $endgroup$



              Hint



              $$2x^2+x(1-5y)+2y^2+y=0$$ is a quadratic equation in $x$



              The discriminant $d=(1-5y)^2-8(2y^2+y)=9y^2-18y+1=(3y-3)^2-8$



              As $d$ needs to be perfect square, $(3y-3)^2-8=z^2$(say)



              $(3y-3-z)(3y-3+z)=8$



              As the multiplicands have the same parity, both must be even



              $dfrac{3y-3-z}2cdotdfrac{3y-3+z}2=2$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 31 '18 at 18:17









              lab bhattacharjeelab bhattacharjee

              226k15157275




              226k15157275












              • $begingroup$
                $$2=(-2)(-1)=2cdot1$$
                $endgroup$
                – lab bhattacharjee
                Dec 31 '18 at 18:21


















              • $begingroup$
                $$2=(-2)(-1)=2cdot1$$
                $endgroup$
                – lab bhattacharjee
                Dec 31 '18 at 18:21
















              $begingroup$
              $$2=(-2)(-1)=2cdot1$$
              $endgroup$
              – lab bhattacharjee
              Dec 31 '18 at 18:21




              $begingroup$
              $$2=(-2)(-1)=2cdot1$$
              $endgroup$
              – lab bhattacharjee
              Dec 31 '18 at 18:21











              4












              $begingroup$

              $$ (2x-y-1)(x-2y+1) = -1 $$
              With integers, you must have either $1 cdot (-1)$ or $(-1) cdot 1$
              Both cases give an intersection of a pair of (non-parallel) lines, so just two points total






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                $$ (2x-y-1)(x-2y+1) = -1 $$
                With integers, you must have either $1 cdot (-1)$ or $(-1) cdot 1$
                Both cases give an intersection of a pair of (non-parallel) lines, so just two points total






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  $$ (2x-y-1)(x-2y+1) = -1 $$
                  With integers, you must have either $1 cdot (-1)$ or $(-1) cdot 1$
                  Both cases give an intersection of a pair of (non-parallel) lines, so just two points total






                  share|cite|improve this answer









                  $endgroup$



                  $$ (2x-y-1)(x-2y+1) = -1 $$
                  With integers, you must have either $1 cdot (-1)$ or $(-1) cdot 1$
                  Both cases give an intersection of a pair of (non-parallel) lines, so just two points total







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 18:29









                  Will JagyWill Jagy

                  103k5102200




                  103k5102200























                      1












                      $begingroup$

                      You have a quadratic equation on $x$ with parameter $y$: $$2x^2+x(1-5y)+2y^2+y=0$$
                      so the discriminat must be prefect square
                      $$ d^2 = (1-5y)^2-8(2y^2+y) = 9y^2-18y+1$$



                      So we have $$8 = 9(y-1)^2 -d^2 = (3y-3-d)(3y-3+d)$$



                      and now should be easy...






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        You have a quadratic equation on $x$ with parameter $y$: $$2x^2+x(1-5y)+2y^2+y=0$$
                        so the discriminat must be prefect square
                        $$ d^2 = (1-5y)^2-8(2y^2+y) = 9y^2-18y+1$$



                        So we have $$8 = 9(y-1)^2 -d^2 = (3y-3-d)(3y-3+d)$$



                        and now should be easy...






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          You have a quadratic equation on $x$ with parameter $y$: $$2x^2+x(1-5y)+2y^2+y=0$$
                          so the discriminat must be prefect square
                          $$ d^2 = (1-5y)^2-8(2y^2+y) = 9y^2-18y+1$$



                          So we have $$8 = 9(y-1)^2 -d^2 = (3y-3-d)(3y-3+d)$$



                          and now should be easy...






                          share|cite|improve this answer









                          $endgroup$



                          You have a quadratic equation on $x$ with parameter $y$: $$2x^2+x(1-5y)+2y^2+y=0$$
                          so the discriminat must be prefect square
                          $$ d^2 = (1-5y)^2-8(2y^2+y) = 9y^2-18y+1$$



                          So we have $$8 = 9(y-1)^2 -d^2 = (3y-3-d)(3y-3+d)$$



                          and now should be easy...







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 31 '18 at 18:20









                          greedoidgreedoid

                          44.8k1156111




                          44.8k1156111






























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