Could I ask something seemingly simple?
Well, let $N$ be a positive odd number (the reason why I set $N$ to be odd is I could actually solve the problem when $N$ is even which is easy) and $a$ is an integer such that $\gcd{(a, N)}=1$ (that is, $a$ has the multiplicative inverse in $\mathbb{Z}_N$.)
I think when $a$ is odd, then $a^{-1} \text{mod} N$ is also odd, and reversely when $a$ is even, then $a^{-1} \text{mod} N$ is also even.
Am I right? As I said, when $N$ is even, it can be easily proved. But when $N$ is odd, it becomes a little confusing.
I expect the answer to be 'Yes' and the proof to not be complicated.
Could any body help me?
Thank you in advance.
When you're working in $\mathbb{Z}_N$ with an odd $N$, there is no well-defined distinction between "even" and "odd" because $2$ is invertible in $\mathbb{Z}_N$, so it divides all elements.
If you take it to mean the parity of the element of $\mathbb{Z}$ in the range $0 \leqslant k < N$ representing the element, then the inverse of an "odd" element can be "even", for example in $\mathbb{Z}_5$, we have $3^{-1} = 2$, and in $\mathbb{Z}_{21}$, we have $11^{-1} = 2$.