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

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Im not sure how to go about it except for trial and error, is there a more efficient way to do this problem?

Feb 9, 2023

#1
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Let's consider the value of i^x for all possible values of x:

i^1 = i i^2 = -1 i^3 = -i i^4 = 1

As we can see, the value of i^x starts repeating after every 4th power. This means that for any positive integer n, the value of i^n can be expressed as i^(n % 4).

Since 1 ≤ x < y ≤ 100, we have four possibilities for the value of (x % 4) and (y % 4):

(x % 4) = 1, (y % 4) = 3

(x % 4) = 2, (y % 4) = 2

(x % 4) = 3, (y % 4) = 1

(x % 4) = 0, (y % 4) = 0

For the first case, we have 25 pairs of (x, y) where x = 1, 5, 9, ..., 97 and y = 3, 7, 11, ..., 99.

For the second case, we have 25 pairs of (x, y) where x = 2, 6, 10, ..., 98 and y = 2, 6, 10, ..., 98.

For the third case, we have 25 pairs of (x, y) where x = 3, 7, 11, ..., 99 and y = 1, 5, 9, ..., 97.

For the fourth case, we have 24 pairs of (x, y) where x = 4, 8, 12, ..., 100 and y = 4, 8, 12, ..., 100.

Therefore, the total number of pairs of integers (x, y) with 1 ≤ x < y ≤ 100 such that i^x + i^y is a real number is 25 + 25 + 25 + 24 = 99.

Feb 9, 2023
#2
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Note that the only way to have a real number is to have both x and y have the same parity (look at the powers of i under guest's answer)

If both x and y are even, there are 50 odd numbers to choose from. Note that if we choose any pair of 2 distinct even numbers, there will be 1 case for x and y. So, in this case, there are \({50 \choose 2} = 1225\) cases.

However, if x and y are odd, we need one to have a remainder of 1 when divided by 4 and the other to have a remainder of 3. There are 25 numbers for x and y, which makes for \(25^2 = 625\) cases.

So, there are a total of \(1225 + 625 = \color{brown}\boxed{1850}\) cases.

Feb 10, 2023