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Asked by vikasg13.hardware | 28 Jun, 2018, 06:10: AM
x2 + 4y2 - 2xy - 2x - 4y - 8 = 0
x2 - 2xy + y2 + 3y2 - 2x - 4y - 8 = 0
(x - y)2 + 3y2 - 2x + 2y - 6y - 8 = 0
(x - y)2 - 2x + 2y + 3y2 - 6y - 8 = 0
(x - y)2 - 2(x - y) + 1 + 3y2 - 6y - 8 - 1 = 0
(x - y - 1)2 + 3y2 - 6y - 9 = 0
(x - y - 1)2 + 3(y2 - 2y - 3) = 0
(x - y - 1)2 + 3(y2 - 2y + 1) - 12 = 0
(x - y - 1)2 + 3(y - 1)2 = 12
Let x - y - 1 = a and y - 1 = b
a2 + b2 = 12
It is possible when (0, 2), (0, -2), (-3, -1), (-3, 1), (3, -1), (3, 1)
Find x and y using these pairs you will get 6 integer pairs.

Answered by Sneha shidid | 29 Jun, 2018, 09:30: AM

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