Compute the number of ordered pairs of integers $(x,y)$ with $1\le x

Question

Compute the number of ordered pairs of integers $(x,y)$ with $1\le x<y\le 100$ such that $i^x+i^y$ is a real number.

anonymous · Answer

$$x=\left\{ 2m,m \in N,1\le m \le 49   \right\}$$
$$y=\left\{ 2n,n \in N,2\le n \le 50 \right\}$$
$$\left( x,y \right)=\left\{( 2m,2n),m<n,m \in N,n \in N,1\le m \le 49,2\le n \le 50 \right\}$$

thomas5267 · Answer

But x=1 and y=3 would also work right?

zmudz · Answer

@surjithayer @thomas5267 , I got (1, 3), (3, 1), (0, 2), and (2, 0) but 4 ordered pairs is wrong :( I can't get the answer unless I give up on the problem, but I really need the points. I'm not sure what @surjithayer 's notation in the set means. Any further help is greatly appreciated!

thomas5267 · Answer

$0\leq x\color{red}{<}y\leq100$

thomas5267 · Answer

$i$ is the imaginary number right?

anonymous · Answer

x and y are even numbers
when m=1,x=2*1=2
when n=2
y=2*2=4
$$\iota ^2=-1$$

thomas5267 · Answer

Nope, x and y being even is not the only cases when $i^x+i^y$ is real.  $-i+i=0\in\mathbb{R},\,1+1=2\in\mathbb{R}$

thomas5267 · Answer

$$
\begin{align*}
x&=2n,\,1\leq n\leq49\
y&=2m,\,2\leq n+1\leq m \leq 50\
n&=1,\,x=2,\,y=4,6,8,\dots,100\
n&=1,\,x=2,\,49\text{ choices of }y\
n&=2,\,x=4,\,y=6,8,\dots,100\
n&=2,\,x=4,\,48\text{ choices of }y
\end{align*}
$$
Choices of $(x,y)$ with $x=2n,\,1\leq n\leq49,\,y=2m,\,2\leq n+1\leq m \leq 50$:
$49+48+47+\cdots+3+2+1=50\times49\div 2=1225$

$$
\begin{align*}
x&=2n-1,\,1\leq n\leq49\
y&=2m-1,\,2\leq n+1\leq m \leq 50\
n&=1,\,x=1,\,y=3,5,7,\dots,99\
n&=1,\,x=1,\,49\text{ choices of }y\
n&=2,\,x=3,\,y=5,7,9,\dots,99\
n&=2,\,x=3,\,48\text{ choices of }y
\end{align*}
$$
Choices of $(x,y)$ with $x=2n-1,\,1\leq n\leq49,\,y=2m-1,\,2\leq n+1\leq m \leq 50$:
$49+48+47+\cdots+3+2+1=50\times49\div 2=1225$

$$
\text{Total} = 1225\times2=2450
$$

thomas5267 · Answer

Something is wrong with my answer.

thomas5267 · Answer

$$
\begin{align*}
x&=4n-3,\,1\leq n\leq 25\
y&=4m-1,\,1\leq n\leq m \leq 25\
n&=1,\,x=1,\,y=3,7,11,\dots,99\
n&=1,\,x=1,\,25\text{ choices of }y\
n&=2,\,x=5,\,y=7,11,15,\dots,99\
n&=2,\,x=3,\,24\text{ choices of }y
\end{align*}
$$
Choices of $(x,y)$ with $x=4n-3,\,1\leq n\leq 25,\,y=4m-1,\,1\leq n\leq m \leq 25$:
$25+24+23+22+\cdots+2+1=26\times25\div2= 325$

$$
\begin{align*}
x&=4n-1,\,1\leq n\leq 24\
y&=4m-3,\,2\leq n+1\leq m \leq 25\
n&=1,\,x=3,\,y=5,9,13,\dots,97\
n&=1,\,x=3,\,24\text{ choices of }y\
n&=2,\,x=7,\,y=9,13,17,\dots,97\
n&=2,\,x=3,\,23\text{ choices of }y
\end{align*}
$$
Choices of $(x,y)$ with $x=4n-1,\,1\leq n\leq 24,\,y=4m-3,\,2\leq n+1\leq m \leq 25$:
$24+23+22+21+\cdots+2+1=25\times24\div2= 300$

Total=$300+325+1225=1850$

This is the answer returned by Mathematica.

zmudz · Answer

1850 is the right answer. The teacher got it by doing casework:

(a) $i^x = i$ and $i^y = -i$, or (b) $i^x = -i$ and $i^y = i$

They both give 625 cases, because there are 25 numbers n for which $i^n = i$ (namely, $n = 1, 4, \ldots, 97$), and there are 25 numbers n for which $i^n = -i$ (namely $n = 3, 7, \ldots, 99$). This gives 1250 cases total.

There are 2500 other cases (50*50) that result from having any pair of even numbers in which it will give a real answer. But then there are overlaps of 50 cases when x=y, so you subtract that. And by symmetry, only half of the remaining will satisfy x<y, so that gives 1850.

thomas5267 · Answer

Basically the same idea but your teacher count all combinations then subtract duplicates where I count all possible combinations directly at a cost of a higher complexity.

zmudz · Answer

yep