<s>[INST] 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!
 [/INST] To get the given solution, combine congruences in your solution with the Chinese remainder theorem: $3\bmod7\land1\bmod11\iff45\mod77$, etc.
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