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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!

J. W. Tanner
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1 Answers1

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

Parcly Taxel
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  • From what I understand, I will get the answers by solving each pair: $x \equiv 3 \pmod 7$ and $x \equiv 1 \pmod {11}$, $x \equiv 3 \pmod 7$ and $x \equiv 10 \pmod {11}$, $x \equiv 4 \pmod 7$ and $x \equiv 1 \pmod {11}$, $x \equiv 4 \pmod 7$ and $x \equiv 10 \pmod {11}$, right? –  Jan 05 '21 at 18:17
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    @fish Yes.${}{}{}$ – Parcly Taxel Jan 05 '21 at 18:24
  • Thanks for your help! –  Jan 05 '21 at 18:26