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Rephrasing the question, it is asking to find the number of solutions to $n|x(x+1)(x-1)$

What I have so far is that if $n$ is prime then the solutions are $x = 0, 1, n-1$, and for $n$ is composite I think I first need to find the number of solutions for when $n$ is a prime power and use the CRT to deduce the final solution. I'm stuck on trying to deduce the number of solutions for when $n$ is a prime power, any hints on how to approach it?

Bill Dubuque
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McBasketball
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    If $ \gcd(n, 2) \neq 2 $, use the fact that $\gcd(x, x+1) = 1, \gcd (x, x-1) = 1, \gcd (x-1, x+1) \leq 2 $ to conclude that either $x, x+1, x-1 \equiv 0 \pmod{n}$. Then deal with $\gcd(n,2) = 2$ separately. – Calvin Lin Oct 28 '20 at 00:55
  • Welcome to Mathematics Stack Exchange. If $p>2$ is prime, $p$ could divide $x$, $x-1$, or $x+1$, but not more than one of those – J. W. Tanner Oct 28 '20 at 00:56
  • Not if $\gcd(x,p^k)=1$ then $x^3 \equiv x \iff x^2 \equiv 1$. Nd if $\gcd(x,p^k)=p^m$ then $x = p^w$. – fleablood Oct 28 '20 at 00:58
  • @CalvinLin, If $n = 15$, $x = 4,5,6$ are all solutions though. – McBasketball Oct 28 '20 at 04:13
  • Since you mentioned you're stuck on how to deduce the number of solutions when $n$ is a prime power, all the above comments apply to prime powers only. You can their hints and CRT to tackle the problem. – player3236 Oct 28 '20 at 04:59
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    @McBasketball My statement was in regards to your point of "When $n$ is a prime power, then use CRT", which is the standard approach in such cases. For prime powers $p^k, p \neq 2$, they contribute $3$ solutions $\mod p^k$. – Calvin Lin Oct 28 '20 at 14:26
  • @CalvinLin Thank you, my misunderstanding. – McBasketball Oct 28 '20 at 19:04
  • For an example using CRT to compute modular roots of polynomials see here. – Bill Dubuque Oct 28 '20 at 22:15

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