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BdMO Online Forum • View topic - $x^2 \equiv x (mod n)$

## $x^2 \equiv x (mod n)$

For discussing Olympiad Level Number Theory problems

### $x^2 \equiv x (mod n)$

Let $n$ be a positive integer. Determine, in terms of n, the number of $x$ such that $x \in {1,2,...n}$ and $x^2 \equiv x(mod n)$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

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dshasan

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Joined: Fri Aug 14, 2015 6:32 pm

### Re: $x^2 \equiv x (mod n)$

If we denote $f(n)$ the desired number, then CRT implies $f(mn)=f(m)f(n)$ whenever $(m;n)=1$.
So it suffices to find $f(n)$ when $n$ is a prime power.
It's quite trivial that there are $2$ solutions for $x$ if $n$ is a prime power. So, the answer is $2^z$ where $z$ is the number of distinct primes in the prime factorization of $n$.
Frankly, my dear, I don't give a damn.

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