Chinese Remainder Theorem: four square roots of 1 modulo N Given an odd composite number $N$, where $N$ is not a prime power, I read the following in a Wikipedia article:

As a consequence of the Chinese remainder theorem, the number $1$ has at
  least four distinct square roots modulo $N$, two of them being $1$ and $-1$.

The square roots of $1$ and $-1$ are obvious to me. What I don't understand is why there are necessarily two others.
Can anyone prove how this result follows from the Chinese remainder theorem?
 A: This follows very simply from the observation that if you have two coprime moduli, $p$ and $q$, then 
$$\begin{cases}
x\equiv a \bmod p \\
x\equiv a \bmod q \\
\end{cases}
\qquad \implies x\equiv a \bmod pq
$$
as a special case of the CRT. (I would like to write the paired equivalence as $x\underset{(p,q)}\equiv (a,a)$)
Then apply this here with 
$$\begin{cases}
x^2\equiv 1 \bmod p \\
x^2\equiv 1 \bmod q \\
\end{cases}
\qquad \implies x^2\equiv 1 \bmod pq
$$
... or $x^2\underset{(p,q)}\equiv (1,1)\implies x^2= 1 \bmod pq$
Then with $p,q>2$ (so that ${-}1{\not\equiv}1$), we can see that $x^2\underset{(p,q)}\equiv (1,1)$ will hold for each of $x\underset{(p,q)}\equiv \{(1,1),$ $(1,-1),(-1,1),$ $(-1,-1)\}$. These will each produce different roots of $1\bmod pq$ with the final values of $x\bmod pq$ determined through the CRT

As an example of how this works out, $21\underset{(5,11)}\equiv (1,-1)$ so $21^2\underset{(5,11)}\equiv (1,1)$ and thus $21^2\equiv 1 \bmod 55$. 
A: If $N$ is divisible by two distinct odd primes (say $p, q$), besides $1, -1$, you can also choose $i$ such that $i \equiv 1 \mod p^{\nu_p(N)}, i \equiv -1 \mod q^{\nu_q(N)}, i \equiv 0 \mod \frac{n}{p^{\nu_p(N)}q^{\nu_q(N)}}$ and $j$ such that $j i \equiv -1 \mod p^{\nu_p(N)}, i \equiv 1 \mod q^{\nu_q(N)}, i \equiv 0 \mod \frac{n}{p^{\nu_p(n)}q^{\nu_q(N)}}$. 
In general if $N$ is divisible by atleast $m$ odd primes, then there are exactly $2^m$ numbers whose square is congruent to $1$ modulo $N$.
Here $\nu_p(m)$ denotes the highest power of $p$ dividing $m$ (for example, $\nu_3(27) = 3$, $\nu_3(18) = 2$, $\nu_3(10) = 0$)
A: If $f:\mathbb Z_N \to 
    \mathbb Z_{p_1^{\alpha_1}} \times Z_{p_2^{\alpha_2}} 
    \times \cdots \times Z_{p_n^{\alpha_n}}$
where $N = p_1^{\alpha_1} \times p_1^{\alpha_1} \times \cdots \times p_n^{\alpha_n}$
is a group isomorphism (via the CRT), then every number of the form
$$f^{-1}(\pm 1, \pm 1, \dots, \pm 1)$$
is a square root of $1$ in $\mathbb Z_N$.
added 9/3/2022
Viewed from the other direction.
If $f: \mathbb Z_{p_1^{\alpha_1}} \times Z_{p_2^{\alpha_2}} 
    \times \cdots \times Z_{p_n^{\alpha_n}} \to \mathbb Z_N$
is a group isomorphism (via the CRT), then every number of the form
$$f(\pm 1, \pm 1, \dots, \pm 1)$$
is a square root of $1$ in $\mathbb Z_N$.
For example, the map $f:Z_3 \times Z_5 \times Z_7 \to Z_{105}$ defined by
$f(x,y,z) = 70x + 21y + 15z \pmod{105}$ is a group isomorphism. We get
$$f( 1, 1, 1) \equiv   1 \pmod{105}$$
$$f( 1, 1,-1) \equiv  76 \pmod{105}$$
$$f( 1,-1, 1) \equiv  64 \pmod{105}$$
$$f( 1,-1,-1) \equiv  34 \pmod{105}$$
$$f(-1, 1, 1) \equiv  71 \pmod{105}$$
$$f(-1, 1,-1) \equiv  41 \pmod{105}$$
$$f(-1,-1, 1) \equiv  29 \pmod{105}$$
$$f(-1,-1,-1) \equiv 104 \pmod{105}$$
and $       1^2 \equiv 76^2 \equiv 64^2 \equiv  34^2 
    \equiv 71^2 \equiv 41^2 \equiv 29^2 \equiv 104^2 \pmod {105}$
A: Let $N = mn$ with $mn$ relatively prime and both odd.
Then among the equivalence classes $\mod N$:
by CRT for each of the following systems of equation there is exactly one solution:


*

*$x \equiv 1 \mod m$ and $x\equiv 1 \mod m$.  (In this case $x \equiv
   1 \mod mn$).

*$y \equiv 1 \mod m$ and $y \equiv -1 \mod m$.

*$w \equiv -1 \mod m$ and $w\equiv 1\mod m$

*$u \equiv -1\mod m$ and $w\equiv -1 \mod m$.(this is $u \equiv -1
   \mod mn$).


These are four distinct values $\mod mn$.
Now all four of these values have the properties that:


*

*$x^2 \equiv y^2 \equiv z^2 \equiv u^2 \equiv 1 \mod m$ and $x^2 \equiv y^2 \equiv z^2 \equiv u^2 \equiv 1 \mod n$.  


By CRT there is exactly one solution $\mod mn$ to that: 
$x^2 \equiv y^2\equiv z^2 \equiv u^2 \equiv 1 \mod mn$.
