Number of lattice points inside a circle 
The number of lattice points inside the circle $x^2+y^2=a^2$ can not be
Options $(a)\; 202\;\;\; (b)\; 203\;\;\; (c)\; 204\;\;\; (d)\; 205$

Try: i have an idea of number of integer points on the circle $x^2+y^2=a^2$
Let $x,y\in\{4n,4n+1,4n+2,4n+3\}$
But no idea how to find number of integer points inside the circle.
Could some help me to solve it , Thanks
 A: Let $C=\{(x,y)\mid x^2+y^2\leq a^2\}$.
The number of lattice points in the $4$ sets $\{(0,y)\in C\mid y>0\}$, $\{(0,y)\in C\mid y<0\}$, $\{(x,0)\in C\mid x>0\}$, $\{(x,0)\in C\mid x<0\}$ is the same.
The number of lattice points in the $4$ sets $\{(x,y)\in C\mid x>0,y>0\}$, $\{(x,y)\in C\mid x>0,y<0\}$, $\{(x,y)\in C\mid x<0,y>0\}$, $\{(x,y)\in C\mid x<0,y<0\}$ is the same.
These sets are disjoint and cover $C$ with exception of lattice point $(0,0)$.
So the number of lattice points in $C$ can be written as $1+4k$ where $k$ is a nonnegative integer.
This rules out the options a),b),c).
A: According to Gauss's circle problem, all choices cannot be ($r$ is radius, $N(r)$ is the number of lattice points):
$$\begin{array}{c|c|c}
r&0&1&2&3&4&5&6&7&8&9&10&11&12\\
\hline
N(r)&1&5&13&29&49&81&113&149&197&253&317&377&441
\end{array}$$
See the graph to verify the numbers $N(8)=197$ and $N(9)=253$:
$\hspace{5cm}$
$$\begin{align}N(8)&=1+(0+1+3+4+7+7+8+11+8)\cdot 4=197\\
N(9)&=1+(0+1+3+4+7+7+8+11+13+9)\cdot 4=253.\end{align}$$
