1x 2 + 6 x + y 2 = 4
Or , ( x + 3 ) 2 = 13 - y 2
For " x " to be an integer , " 13 - y 2 " must be a perfect square , which can happen only for -
y = 2 , - 2 , 3 , - 3 ................ Accordingly , " x " will also take different values .
Hence , there are 4 ordered pairs of ( x , y ) .
1then there will be 8 solns..pls confirm
-1,2
-1,-2
-5,2
-5,-2
and 4 more with y=+ -3
1sorry replace 2 by 3 in all above ones,,
similarly there'll be 4 more with y= +2,- 2
1All the solutions are -
( - 1 , 3 ) ; ( - 1 , - 3 ) ; ( - 5 , 3 ) ; ( - 5 , - 3 ) ; ( 0 , 2 ) ; ( 0 , - 2 ) ; ( - 6 , - 2 ) , ( - 6 , - 2 ) ;
Hence , there are 8 solutions .
EDIT - Thanks to " Seoni " , I corrected a major blunder . And Seoni , thanks to you mate for giving me a question worth trying .
