Ordered pairs $(x,y)$ that satisfy the equation $x^2+y^2 = 2013$ (1) The no. of Integer ordered pair $(x,y)$ that satisfy the equation $x^2+y^2 = 2013$
(2) The no. of Integer ordered pair $(x,y)$ that satisfy the equation $x^2+y^2 = 2014$
My solution for fisrt::
Here R.H.S is a odd quantity means L.H.S must be odd  which is possibly only when one square 
quantity is even and other is odd 
So Let $x=2r$ and $y=2s+1$ and equation is $4(r^2+s^2+s)+1=2013$
Now How can I calculate after that
Thanks

Note that $2013=3\cdot11\cdot61, 2014=2\cdot19\cdot53$. 
$x^2+y^2=2013 \Rightarrow 3 \mid x^2+y^2 \Rightarrow 3\mid x, y \Rightarrow 9 \mid x^2+y^2=2013$, a contradiction.
$x^2+y^2=2014 \Rightarrow 19 \mid x^2+y^2 \Rightarrow 19\mid x, y \Rightarrow 19^2 \mid x^2+y^2=2014$, a contradiction.
Thus there are no integer solutions.
Here we have used the fact that $-1$ is not a quadratic residue $\pmod{p}$ if $p \equiv 3 \pmod{4}$.
