How many involutions does $\Bbb Z/999$ have?

Question

How many involutions does $\Bbb Z/999$ have?

Order of $g$ is donate by $o(g)$. And $o(g)=2$ iff $g$ is an involution.

I know that order of $\Bbb Z/999$ is $999$ but how can I check $999$ elements one by one to find the answer.

So could you give me some hint to do it?

Answer 1

Suppose $g\in \Bbb Z_{999}$ has order two. Then, by Lagrange's Theorem, $$2=\lvert \langle g\rangle\rvert\mid \lvert \Bbb Z_{999}\rvert=999,$$ a contradiction; hence $\Bbb Z_{999}$ has no elements of order two.