<s>[INST] Chinese Remainder Theorem RSA I want to solve the following modular quadratic equation:
$x^2 \equiv 188 \pmod {437}$ using the fact that $437$ can be factorized by the primes as: $19⋅23$.
So far I have done:
$$x^2 \equiv 188 \pmod{19} \Rightarrow x^2 \equiv 17 \pmod {19}\\
x^2 \equiv 188 \pmod {23} \Rightarrow x^2 \equiv 4 \pmod {23}$$
$$x^2 \equiv 7 \pmod{ 19} \Rightarrow x \equiv \pm 6 \pmod {19}\\
x^2 \equiv 4 \pmod {23} \Rightarrow x \equiv \pm 2 \pmod {23}$$
And the last step. But I do not understand how to combine:
$$x \equiv  6 \pmod {19} \text{ and } x \equiv  2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv  2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod{ 437}\\
x \equiv  6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}$$
Please help me fill the "insert-her".
 [/INST] You can find the solution using this calculator just fill in $$ x=6 \mod 19$$ and $$
x=2 \mod 23
$$
Your solution is $25$ and indeed $188 \equiv 25^2 \pmod {437} $
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