Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$ Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$

Using the Chinese Remainder Theorem, I obtained $x^2 \equiv 23 \pmod{7}$ and $x^2 \equiv 23 \pmod{11}$.
For $x^2 \equiv 23 \pmod{7}$, the answers are:
\begin{align}
x \equiv 3 \pmod{7} \\
x \equiv 4 \pmod{7}
\end{align}
For $x^2 \equiv 23 \pmod{11}$, the answers are:
\begin{align}
x \equiv 1 \pmod{11} \\
x \equiv 10 \pmod{11}
\end{align}
But the solutions for $x^2 \equiv 23 \pmod{77}$ are:
\begin{align}
x \equiv 10 \pmod{77} \\
x \equiv 32 \pmod{77} \\
x \equiv 45 \pmod{77} \\
x \equiv 67 \pmod{77}
\end{align}
Why are my answers not the same as the solutions? Can someone help me with this? Thanks!

To get the given solution, combine congruences in your solution with the Chinese remainder theorem: $3\bmod7\land1\bmod11\iff45\mod77$, etc.
