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I'm working on Shor's algorithm. In Wiki, there says that if N is the product of two coprime integers greater than 2. Then, exist at least four distinct square root of 1 (mod N). And, I know I could reduce this congruence relation into two congruence relations. Then, we could apply Chinese Remainder theorem. But, I don't know how to apply this theorem to this statement. May someone give me some hints? Thank you!!

Andrew
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  • Try it with an example. Use the CRT to find the four square roots of $1$ $\pmod {15}$. That one is small enough so you can do it by hand as well. – lulu Feb 14 '20 at 15:21
  • In order to use the Chinese remainder theorem you'd need the factors of $N$, which you do not have. Suppose you have two different roots of unity mod $N$, $a$ and $b$ such that $a^2=b^2=1$. Then $a^2-b^2=0=(a-b)(a+b)$ and if you are lucky, you found a zero-divisor in $N$ i.e. $a-b\neq0$ and / or $a+b\neq0$ hence $\gcd(N,a-b)\neq1$ and you thus found a divisor of $N$. – emacs drives me nuts Feb 14 '20 at 15:21
  • I have tried the small case. But, I still can't find out the reason why at least four distinct roots. Besides, why $gcd(N, a-b)\neq 1$? I can't get it. – Andrew Feb 14 '20 at 15:52
  • See the answers in the linked dupes for how to use CRT to lift roots of polynomials. In particular my answers explicitly describe how it applies in general. – Bill Dubuque Feb 14 '20 at 16:36

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