Square root of 1 with mod how to prove it?

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SUMMARY

The discussion centers on proving that if \( n \) is an odd integer with \( k \) distinct prime factors, then the equation \( x^2 \equiv 1 \mod n \) has exactly \( 2^k \) solutions. The participants suggest using the Chinese Remainder Theorem (CRT) to combine results from individual prime powers \( p^k \) to establish the proof. This approach allows for a structured method to derive the number of roots without resorting to the generalized form \( x^2 \equiv a \mod n \).

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  • Understanding of modular arithmetic, specifically \( x^2 \equiv a \mod n \)
  • Familiarity with the Chinese Remainder Theorem (CRT)
  • Knowledge of prime factorization and distinct prime factors
  • Basic concepts of number theory
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Mathematicians, number theorists, and students studying modular arithmetic and number theory who seek to understand the properties of quadratic residues and their proofs.

jacquelinek
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I know that if n is odd and has k distinct prime factors, then the number of roots, x^2 = 1 (mod n), is equal to 2^k.
However, I don't know how to give a formal proof to it.
I simply want to bypass the generalized form x^2 = a (mod n).
How can I prove it directly?
Thank you.
 
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Hmm... assuming it holds mod p^k, you could just CRT the results together.
 

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