What $x$ satisfies $x^2 \equiv x \pmod{n}$ for arbitrary $n \in \mathbb{N}$?

Asked May 23 '20 at 18:33

Active May 23 '20 at 18:54

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$\textbf{Question}$: Let $n$ be a positive integer. Determine, in terms of $n$, the number of $x$ where $x \in {1,2,\ldots,n}$ for which $x^2 ≡ x \pmod n.$

$\textbf{My solution}$: Let $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$. Now for each of the prime power we have two choices :

$p_i^{a_i} \mid x$, $\; $or,
$p_i^{a_i} \mid x-1$.

Then Chinese Remainder Theorem would make sure that these $n$ congruences result in one unique $x.$

Hence, our answer is $2^{\omega(n)}$ where $\omega(n)$ denotes the number of distinct prime factors of n.

I wanted to verify my answer and I would also appreciate any kind of remark.

edited May 23 '20 at 18:47

saru

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asked May 23 '20 at 18:33

Yes it's me

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Yes, same as in the linked dupe. – Bill Dubuque May 23 '20 at 18:49
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Anyway, your argument is correct. – Jyrki Lahtonen May 23 '20 at 18:53
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@Jyrki I'll add that too (the general case is handled in the first link too, though less explicitly, but more generally, i.e. any polynomial $\in\Bbb Z[x],,$ not only $,x^2-x)$ – Bill Dubuque May 23 '20 at 18:54
Thanks @BillDubuque. I agree that it is a good idea to have a link to a case where polynomials are also discussed. Deleting the obsolete comment. – Jyrki Lahtonen May 23 '20 at 18:56
@Jyrki Yes, the more the merrier when it comes to good dupe links. – Bill Dubuque May 23 '20 at 18:57

What $x$ satisfies $x^2 \equiv x \pmod{n}$ for arbitrary $n \in \mathbb{N}$?

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