# Knowing prime factorization of $n$ determine number of solutions to $x^2 \equiv 1\ (\textrm{mod}\ n)$ [closed]

How to determine number of solutions to $x^2 \equiv 1\ (\textrm{mod}\ n)$, when I know prime factorization of $n$?

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## closed as off-topic by Bruno Joyal, user1337, Hagen von Eitzen, Jyrki Lahtonen♦, T. Bongers Nov 5 '13 at 22:50

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I assume the first step is to rewrite this as $n|(x+1)(x-1)$. – Mike Nov 2 '13 at 14:39
Hi @Machiawliczny, could you please give some more context? For example, do you know about the Chinese Remainder Theorem? – Johannes Kloos Nov 2 '13 at 18:01
possible duplicate of Roots of $x^2 -1$ in $\mathbb{Z}\_m$ with $m$ not a prime number. To tell you the truth I was surprised not to find an older dup. – Jyrki Lahtonen Nov 3 '13 at 10:19

Assume $n=2^ap_1^{a_1}\dots p_m^{a_m}$. Then the number of solutions are $2^m$ if $a=0$ or if $a=1$; $2^{m+1}$ if $a=2$ and $2^{m+2}$ if $a\geq 3$.

Proving this takes a chapter in Vinogradov's book.

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Let me see if I can expand on my comment and try to prove Ivan's answer. As I said in the comment, if

$$x^2\equiv1\pmod n$$ $$x^2-1=(x+1)(x-1)\equiv0 \pmod n$$ $$n|(x+1)(x-1)$$

For each prime power factor $p_m^{a_m},p\ne2$, either $p_m^{a_m}|x-1$ or $p_m^{a_m}|x+1$. If $n$ is odd, there are $2$ choices for each prime power. Each set of choices gives $1$ solution $\pmod n$ by the Chinese remainder theorem, so there are $2^m$ possibilities.

$2$ is different since $x-1\equiv x+1\pmod 2$. If $2^a|p$ and $a=1, x\equiv 1\pmod2$ is our only choice. If $a=2$, then either $x-1$ or $x+1$ is divisible by $4$, so $2$ choices for $x\pmod4$. If $a>2$, since $\gcd(x-1,x+1)|2$ we have either

$$2^{a-1}|x-1$$ $$x\equiv1\pmod{2^{a-1}}$$ $$x\equiv1,2^{a-1}+1\pmod{2^a}$$ $$\text{or}$$ $$2^{a-1}|x+1$$ $$x\equiv-1\pmod{2^{a-1}}$$ $$x\equiv-1,2^{a-1}-1\pmod{2^a}$$

so $4$ choices for $x\pmod{2^a}$.

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All primes are odd except 2, which is the oddest of all. – D.E.Knuth. – lhf Dec 10 '13 at 4:43