Answer: $A$
Remainder when $2011$ is divided by $4 = 3$
Now, we know that the square of any even number is divisible by $4$.
$$\implies (2k)^2 = 4 \times k^2,\;where\;k\in Even\;number$$
Also, square of any odd number gives the remainder $1$ when divided by $4$:
$$\implies (2k+1)^2 = 4k^2 + 4k + 1= 4 \times (k^2 + k ) + 1$$
Thus the square of any integer when divided by $4$ either gives $0$ or $1$ as the remainder.
Now, when we add $x^2 + y^2 $ together.
Possible remainders $ = 0+0 = 0, \;0+1= 1, \;1 + 0 = 1, \; 1 + 1 = 2 = \bf\{0, 1, 2\}$
So, we will never get $3$ as the remainder.
Hence, the number of integer solutions for the above equation $=0$
$\therefore A$ is the right option.