How to show that a the group $( \mathbb{Z} / n \mathbb{Z} )^* $ has elements of order $2$

Question

Let $ n \in \mathbb{N} $ have prime factorization $ n = 2^e p_1^{e_1}...p_r^{e_r}, $

with $ 0 \leq e $ and pairwise odd primes $ p_1,...,p_r $

and $ 1 \leq e_1,...,e_r. $

How do you prove that the group $ ( \mathbb{Z} / n \mathbb{Z})^* $ then has

$ \left\{ \begin{array} {ll} 2^r-1, \text{ if } e=0 \lor 1 \\ 2^{r+1} -1 , \text{ if } e=2 \\ 2^{r+2} -1 , \text{ if }3 \leq e \end{array} \right.$

elements of order $2$?

Appreciate any help, because I am not really sure how to prove that !

yes, but i can't see how to solve it with the chinese remainder theorem — constant94, Dec 17 '19 at 16:36
consider solutions to $x^2\equiv1$ in $(\mathbb Z/2^e\mathbb Z)^$, $(\mathbb Z/p_1^{e_1}\mathbb Z)^$, ..., $(\mathbb Z/p_r^{e^r}\mathbb Z)^*$ — J. W. Tanner, Dec 17 '19 at 17:21

score 0 · Answer 1 · answered Dec 17 '19 at 20:43

By the Chinese remainder theorem, solutions of $a^2\equiv1\bmod n$

correspond to solutions of $a^2\equiv1 \bmod 2^e$, $a^2\equiv1 \bmod p_1^{e_1},\dots,$ and $a^2\equiv1 \bmod p_r^{e_r}$.

If $e=0$ or $1$, there is one solution to $a^2\equiv1\bmod2^e: \;a\equiv1 \bmod 2^e$.
If $e=2$, there are $\color{blue}{2}$ solutions to $a^2\equiv1\bmod2^e:\; a\equiv1$ or $-1\bmod 2^e$.
If $e\ge3$, there are $\color{brown}4$ solutions to $a^2\equiv1\bmod 2^e:\; a\equiv\pm1$ or $2^{e-1}\pm1\bmod 2^e$.
There are two solutions to $a^2\equiv1\mod p_i^{e_i}:\; a\equiv1$ or $-1\bmod p_i^{e^i}$.

Therefore, putting these together by the Chinese remainder theorem,

there are $2^r$ solutions to $a^2\equiv1\bmod n$ if $e=0$ or $1$,

$\color{blue}2\times2^r=2^{r+1}$ solutions to $a^2\equiv1\bmod n$ if $e=2$,

and $\color{brown}4\times2^r=2^{r+2}$ solutions to $a^2\equiv1\bmod n$ if $e\ge3$.

In each case, one of the solutions is $a\equiv1\bmod n$, which has order $1$, and the others have order $2$.

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