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1 solution
Rewrite the following equation as: 2 x 2 + 8 y 2 − 4 x y − 4 x − 8 y − 1 6 = 0
After that, it can be written as: ( x − 2 ) 2 + ( x − 2 y ) 2 + ( 2 y − 2 ) 2 = 2 4
Now, assume that 2 4 = a 2 + b 2 + c 2 with a ≥ b ≥ c . Then 4 ≥ a ≥ 3 otherwise the sum gets too high or low. Applying a = 3 , we get b 2 + c 2 = 1 5 . But 15 is of the form 4 m + 3 . Therefore, no solution. Applying a = 4 , we get b 2 + c 2 = 8 . We see that b = c = 2 . Therefore, ( a , b , c ) = ( 4 , 2 , 2 ) , ( − 4 , − 2 , − 2 ) , ( − 4 , 2 , 2 ) , ( − 4 , − 2 , 2 ) , ( 4 , − 2 , − 2 ) , ( 4 , 2 , − 2 ) .
Now, we have various cases. Doing some amount of casework will give you the following pairs x , y : ( − 2 , 0 ) , ( 0 , − 1 ) , ( 0 , 2 ) , ( 4 , 0 ) , ( 4 , 3 ) , ( 6 , 2 )