The number of elements of order 2 in $U(\mathbb{Z}_{p^2q})$ where $p,q$ are distinct primes Let $p,q$ be distinct primes. Let's consider the multiplicative group of the units of $\mathbb{Z}_{p^2q}$, namely $U(\mathbb{Z}_{p^2q})$.

How many elements of order $2$ does $U(\mathbb{Z}_{p^2q})$ have?


My attempt:
$p^{2}q|(a-1)(a+1)$.
Now the possibley combinations are 
i) $p^{2}|(a-1)$ and $q|(a+1)$ so $a=p^2+1 $ and $a=(q-1)$. 
ii) $p^{2}|(a+1)$ and $q|(a-1)$. So $a = p^2-1 $ and $a=(q+1)$.
iii)$ p|(a-1)$ and $pq|(a+1)$ so $a= (p+1)$ and $a=pq-1$
iv) $p|(a+1)$ and $pq|(a-1)$ so $a =p-1$ and $a =pq+1$
v)  $p^2q|(a+1)$ so $a =p^2q-1$
This has been my attempt.
 A: There is one element of order $2$ which you missed: 
$p^2q|(a-1)$ and so $a=1$. 
Otherwise your attempt is OK.
A: If $p$ and $q$ are distinct primes, then:
$$U(\mathbb{Z}_{p^2q})\cong U(\mathbb{Z}_{p^2})\times U(\mathbb{Z}_{q}) \tag 1$$
So, your problem boils down to finding the elements of order $2$ of $U(\mathbb{Z}_{p^2})$ and $U(\mathbb{Z}_{q})$.


*

*Let's first assume $p,q$ odd primes. Then, both $U(\mathbb{Z}_{p^2})$ and $U(\mathbb{Z}_{q})$ are cyclic of even order; it follows that they both have $\varphi(2)=1$ element of order $2$ ($\varphi$ is Euler's function), say $a$ and $b$ respectively:


therefore, if $p,q$ are distinct odd primes, then $U(\mathbb{Z}_{p^2})\times U(\mathbb{Z}_{q})$ has three elements of order $2$, namely $(a,1)$, $(1,b)$ and $(a,b)$; by $(1)$, so has $U(\mathbb{Z}_{p^2q})$.

We are left with the two cases $(p,q)=(2,odd\space prime)$ and $(p,q)=(odd\space prime,2)$:


*If $(p,q)=(2,odd\space prime)$, then $U(\mathbb{Z}_{p^2})\times U(\mathbb{Z}_{q})=U(\mathbb{Z}_{4})\times U(\mathbb{Z}_{q})=\{1,3\}\times U(\mathbb{Z}_{q})$; let $b$ the only element of order $2$ of $U(\mathbb{Z}_{q})$; we have that:

the elements of order $2$ of $U(\mathbb{Z}_{4})\times U(\mathbb{Z}_{q})$ are: $(1,b)$, $(3,1)$ and $(3,b)$. So, $U(\mathbb{Z}_{4q})$ has three elements of order $2$.


*If $(p,q)=(odd\space prime,2)$, then $U(\mathbb{Z}_{p^2})\times U(\mathbb{Z}_{q})=U(\mathbb{Z}_{p^2})\times U(\mathbb{Z}_{2})=U(\mathbb{Z}_{p^2})\times \{1\}$; let $a$ the only element of order $2$ of $U(\mathbb{Z}_{p^2})$; we have that:

the only element of order $2$ of $U(\mathbb{Z}_{p^2})\times U(\mathbb{Z}_{2})$ is $(a,1)$. So, $U(\mathbb{Z}_{2p^2})$ has just one element of order $2$.

This completes the survey for $U(\mathbb{Z}_n)$ in the case $n=p^2q, p\ne q$.
