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Find all integers x, y satisfying (x-y)^2+2y^2=27

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Solution:  We have ,  (x-y)^2+2y^2=27

              (x-y)^2,2y^2geq 0  and since 2y^2 is even  (x-y)^2

        is odd and hence   (x-y)  should be odd so the different 

      possibilities for (x-y)^2 and y^2 are (1,13),(9,9),(25,1).

     Corresponding  to y^2=13 , there is no solution that y is 

    an integers so taking the other two pairs , we get 

                        x-y=pm3 ,y=pm3 ..........(1)

                       x-y=pm5,y=pm1   ..........(2)

     Sloving Eq . (1) we get 

                       (0,3),(6,3),(0,-3),(-6,-3)

  Sloving Eq. (2) we get

                     (6,1),(-4,1),(-6,-1),(4,-1).

 

 

Posted by

Deependra Verma

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