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
 A: 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 $p \equiv 3 \pmod{4}$.
A: If you want a very elementary approach, here's one you can do by hand:
Observe that squares of odd numbers end with $1, 5, 9$, and those of even numbers end with $0, 4, 6$. When adding, the lowest significant digit in 2013 is $3$, which from the above cases, can only be obtained by adding two squares one of which ends with a $9$, and the other which ends with a $4$. Then, you can see that the possible values of odd numbers can only be: $3, 7, 13, 17, 23, 27, 33, 37, 43$. Now notice that $2013 - \text{ square of these odd numbers }$ must also be a perfect square. But you get $2004, 1964, 1844, 1724, 1484, 1284, 924, 644, 164$, none of which is a perfect square. So, there are no possible solutions.
Repeat the same for 2014.
A: Hint $\ $ Apply the theorem below for $\rm\ p,n \,=\, 3,2013;\,\ 19,2014.$ 
Thoeorem $\ $ If  $\rm\  n = x^2\! + y^2\ $ and prime $\rm\ p\mid n,\ p^2\nmid n,\ $ then $\rm\: p \ne 3+4k.\:$ 
Proof $\ $ Deny, so $\rm\:p = 3+4k.\:$ If $\rm\,p\mid x\,$ then $\rm\,p\mid x,\:x^2\!+y^2\:\Rightarrow\:p\mid y^2\:\Rightarrow\:p\mid y,\:$ by $\rm\,p\,$  prime, thus $\rm\,p\mid x,y\:\Rightarrow\:p^2\mid x^2\!+y^2,\,$ contra hypothesis. Hence $\rm\,p\nmid x,\:$ and, by symmetry, $\rm\,p\nmid y,\:$ therefore
$$\rm mod\ p\!:\,\ x,y\not\equiv 0,\ \ x^2\!+y^2 \equiv 0\,\ \Rightarrow\ {-}y^2 \equiv x^2\ \Rightarrow\ \color{#C00}{{-}1}\equiv x^2/y^2 = \color{#C00}{(x/y)^2}\:$$ 
$$\rm\Rightarrow\  {-}1 = (\color{#C00}{-1})^{1+2k}\!\equiv(\color{#C00}{(x/y)^2})^{1+2k}\!\equiv (x/y)^{\,p-1}\!\equiv 1\:\Rightarrow\:2\equiv 0\:\Rightarrow\:p\mid 2,\ \ contra\ \ p\ odd$$
A: Since $3\mid 2013$, and $-1$ is a quadratic non-residue of $3$, one of $x$ and $y$, hence both $x$ and $y$ must be divisible by $3$. But then this tells us that $3^2\mid 2013$, a cnotradiction. So for (1) there is no solution.
Similarly, for (2), $-1$ is a non-residue of $19$ and so $19$ must divide both $x$ and $y$, again a contradiction.
Therefore the answers for both questions are $0$.  
