i^x+i^y... (x.y) pairs Question :" Find the number of ordered pairs $(x,y)$ with $1 \le x < y \le 100$ and $i^x+i^y$ is real."
My work so far: 
I tried using casework on $x$ when taken mod 4 to get the following cases
$x \equiv 1 \pmod 4$, I got a total of 325 $(x,y)$ pairs.
$x \equiv 3 \pmod 4$, I got a total of 276 $(x,y)$ pairs.
$x \equiv 2,4 \pmod 4$, I got a total of 1225 $(x,y)$ pairs.
Can anybody check that my numbers are right?...

Since $i^x$ is either $i,-1,-i,$ or 1, $i^x+i^y$ will be real when either a) x and y are both even, or b) x and y are both odd, with $x\not\equiv y \ (\mod 4)$.
In the first case, there are $50\choose 2$ ways to select x and y; and in the second case, there are $25\times25$ ways to choose x and y, since there are 25 integers in $\{1,\cdots,100\}$ which are congruent to 1 (or respectively, to 3)
mod 4.  Therefore there should be 1850 such pairs.
I believe you should have 300 pairs when $x \equiv 3 \pmod 4$.
