Imaginary Counting

The given expression can be written as:

${ i }^{ a+1 }+{ i }^{ b+1 }=2$

As we know from imaginary unit powers, ${ i }^{ k }$ (with $k$ positive integer) can only be

$i,-i,-1\quad or\quad 1$

Then, the only solution for the expression ${ i }^{ a+1 }+{ i }^{ b+1 }=2$

is ${ i }^{ a+1 }={ i }^{ b+1 }=1$

Yet from imaginary unit powers, ${ i }^{ k }=1$ if and only if k is multiple of 4. Then, we can write $a+1=4n$ Where $n$ is positive integer. Couting all possible values for $1\le a\le 100$ , we have $25$ possible values for $a$ . Similarly, we have $25$ values for $b$ .

We want the number of ordered pairs $(a,b)$ . Cartesian Product (Rule of Product):

$25\quad possibilities\quad for\quad a\quad 25\quad possibilities\quad for\quad b\Rightarrow 25X25=625$ ordered pairs.