$x^a-x^b$ is odd function, and $1 \leq a,b, \leq 100$

Asked Sep 06 '20 at 16:25

Active Sep 06 '20 at 16:32

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How many ordered pairs of positive integers $(a, b)$ with $1 \le a, b \le 100$ exist such that the function $f(x) = x^a - x^b$ is an odd function.

I know that a function is only odd iff both $a$ and $b$ are odd, and there are $50$ odd numbers in between $1$ and $100$ inclusive. Therefore, I think that the answer is 2500 (this includes the case where $a=b$ because $f(x)=0$ is also an odd function). However, this is wrong. I do not know what I am doing wrong, possibly with the restriction. Can anyone identity my error?

edited Sep 06 '20 at 16:32

V.G

4,196

asked Sep 06 '20 at 16:25

If $a=b$ are even, the given function is odd. – TheSilverDoe Sep 06 '20 at 16:34
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What about the cases where $a=b$ and both are even? – Sep 06 '20 at 16:34
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Ahh, thank you for the catch. So it should be 2500+50=2550. – Sep 06 '20 at 16:38

$x^a-x^b$ is odd function, and $1 \leq a,b, \leq 100$

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