a) Ta có f(x) = (x+2)(x-2)q(x) + 10 = (x^2 - 4)q(x) + 10
f(x) = (x-4)(-5) + r = -5x + 20 + r
Ta suy ra r = 10
Vậy f(x) = x^2 - 4 + 10 = x^2 + 6

b) Ta có D = n - n + 2 = 2
Để D là số chính phương, ta cần giải phương trình x^2 = 2
Suy ra n = 2

c) Phương trình đã cho là x + xy - 2012x - 2013y - 2014 = 0
Simplifying the equation, we get:
x(1-y) - 2012(x+1) = 2014
x(1-y) - 2012x - 2012 = 2014
x(1-y-2012) = 4026
x(1-y-2012) = 2*3*11*61
Since x is an integer, 1-y-2012 must be a factor of 4026.
Factors of 4026 are 1, 2, 3, 6, 11, 22, 33, 61, 66, 122, 183, 366, 2013, 4026
Let's try each factor:
1-y-2012 = 1 -> y = -2012
1-y-2012 = 2 -> y = -2013
1-y-2012 = 3 -> y = -2014
1-y-2012 = 6 -> y = -2017
1-y-2012 = 11 -> y = -2022
1-y-2012 = 22 -> y = -2033
1-y-2012 = 33 -> y = -2044
1-y-2012 = 61 -> y = -2060
1-y-2012 = 66 -> y = -2065
1-y-2012 = 122 -> y = -2133
1-y-2012 = 183 -> y = -2194
1-y-2012 = 366 -> y = -2377
1-y-2012 = 2013 -> y = 1
1-y-2012 = 4026 -> y = 2015

Therefore, the integral solutions to the equation are (x, y) = (2015, 1)