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I'm trying to find out how can you figure the number of solutions for a quadratic congruence in a composite modulo. I've seen some similar questions in this website but I didn't understand the answers in any of them, would appericiate anyone trying to explain for me.

thank you!

user670028
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    What examples might you be struggling with ? –  Jul 30 '19 at 21:06
  • For example this previous Question seems to be a more specific version of what you've asked. If you find the discussion there incomprehensible, it should give you the opportunity to frame your Question around the points where you feel lost. – hardmath Jul 30 '19 at 21:20
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    I'm looking for a general formula, I think the user 'Theyaoster' in the previous question tried to give something similar towards the end of his comment but I couldn't figure out what he means, this is the quote:

    "As for how many solutions the congruence has, by the Chinese Remainder Theorem, it would be the product over the number of solutions each of x2≡a(modpaii) has."

     – user670028 Jul 30 '19 at 22:49
  • The idea is ultimately you piece together counting the number of solutions modulo composite $m$ by multiplying the counts of solutions modulo the prime power factors of $m$. In the simplest case $m=pq$ is a product of two distinct primes. Multiply the number of solutions mod $p$ by the number of them mod $q$ and you will have the number of solutions mod $m$. – hardmath Jul 31 '19 at 01:59
  • Got it! thank you very much. I was wondering now how do we make sure none of those solutions are congruent? I mean if we build a solution for mod $n$ from solutions mod $p$ and mod $q$ through Chinese Remainder Theorem, how do we make sure they are not congruent in composite mod $n$ and therefore less than the actual multiplication of number of solutions in moduli of the prime factors? – user670028 Jul 31 '19 at 06:03
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    If two solutions are different mod $p$ (say), then they are also different mod $m=pq$. The situation is similar if the composite modulus $m$ has more prime divisors or some prime power divisors. The most interesting part of the subject involves Hensel lifting to construct solutions modulo a prime power $p^n$ from ones modulo $p$. – hardmath Jul 31 '19 at 20:13

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