Square roots modulo $pq$ where $p$, $q$ are distinct primes - confusion with quote I have a doubt about the following quote from a book:

The Chinese Remainder Theorem implies that, if $p$ and $q$ are
  distinct primes, then $s$ is a square modulo $pq$ if and only if $s$
  is a square modulo $p$ and $s$ is a square modulo $q$. In particular,
  if $s \equiv x^2 \equiv (x')^2 \pmod p$ where $x \neq x'$, and
  likewise $x \equiv y^2 \equiv (y')^2 \pmod q$, then $s$ has exactly
  four square roots modulo $pq$, namely,
$$s \equiv (xy)^2 \equiv (x'y)^2 \equiv (xy')^2 \equiv (x'y')^2 \pmod{pq}$$

As an attempt to understand the quote, suppose I want to find the square roots of $11$ modulo $133$. So, I have $s = 11$, and, since $133 = 7 \times 19$, I have $p = 7$ and $q = 19$. I first need to separately find the roots of $11$ modulo $7$ and of $11$ modulo $19$:


*

*To find the roots modulo $7$, solve $x^2 \equiv 11 \pmod 7$. Since $11 \equiv 2^2 \equiv 5^2 \pmod{7}$, the roots are $x = 2$ and $x' = 5$.

*To find the roots modulo $19$, solve $x^2 \equiv 11 \pmod{19}$. Since $11 \equiv 7^2 \equiv 12^2 \pmod{19}$, the roots are $y = 7$ and $y' = 12$.


The quote is saying that $xy$, $x'y$, $xy'$ and $x'y'$ are roots of $s$ modulo $pq$. However, this seems to be untrue in my example:


*

*$xy = 2 \times 7 = 14$, but $11 \not\equiv 14^2 \pmod{133} $.

*$xy' = 2 \times 12 = 24$, but $11 \not\equiv 24^2 \pmod{133} $.

*$x'y = 5 \times 7 = 35$, but $11 \not\equiv 35^2 \pmod{133} $

*$x'y' = 5 \times 12 = 60$, but $11 \not\equiv 60^2 \pmod{133} $
Am I missing something?

The Chinese remainder theorem must be applied correctly.
In your example, it says that there is exactly one $a \pmod{133}$ such that $a \equiv 2 \pmod{7}$ and $a \equiv 7 \pmod{19}$. See, for example, here.
Since we have $1 = 3 * 19 - 8 * 7$, we put $a = 2 * 3 * 19 - 7 * 8 * 7 = -278$ and find that, indeed $a^2 \equiv 11 \pmod{133}$. An explanation of this calculation, can be found, for example, here.
Similar calculations for the other combinations of roots of $11$ mod $7$ and $19$ produce the following four distinct roots of $11$ mod $133$.
$$a \equiv 121 \pmod{133}$$
$$b \equiv 107 \pmod{133}$$
$$ c \equiv 26 \pmod{133}$$
$$ d \equiv 12 \pmod{133}$$
