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Question: Let $n \in \mathbb N$ not a prime power and not twice a prime power. Let $s \in \mathbb Z_n$ such that $s^2 \equiv 1 \pmod n$ but $s \not\equiv \pm1 \pmod n$. Is it true that always there exists a factorization $n = n_1 n_2$ with $n_1, n_2 > 1$, $(n_1,n_2)=1$, $s \equiv -1 \pmod {n_1}$ and $s \equiv 1 \pmod {n_2}$?

My partial answer: If there exist, the numbers $n_1,n_2$ satisfy $(n_1,n_2)\in\{1,2\}$. In fact, if $d \mid (n_1,n_2)$ then $d \mid s+1$ and $d \mid s-1$, so $d\in\{1,2\}$. Let $n_1=(s+1,n)$ and $n_2 = n/n_1$. If $(n_1,n_2) = 1$, then we are done. Otherwise, suppose that $(n_1,n_2)=2$ and let $n_1 = 2m_1$ and $n_2 = 2m_2$, where $(m_1,m_2)=1$. The latter implies, for example, that $4 \mid n$, $s \equiv 1 \pmod 4$, and either $m_1$ or $m_2$ is odd. Afterwards, I considered the parities of $m_1$ and $m_2$. I tried to check if the factorization $n = t_1t_2$ works, where $t_1 = (\frac{s+1}{2},n)$ and $t_2 = n/t_2$, but it will depend on other parities and I wasn't able to complete all the cases.

Can someone provide a hint that works?

Sávio
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$n$ factors as a product of prime powers. Prove that $s$ is either $1$ or $-1$ mod each prime power factor. Let $n_1$ be the product of prime power factors for which $s$ is $-1$, and let $n_2$ be the product of prime power factors for which $s$ is $1$.

vujazzman
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    How do you ensure that $n_1>1$ and $n_2>1$? – lhf Apr 07 '20 at 12:26
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    $s\ne \pm 1 \pmod n$ – vujazzman Apr 07 '20 at 18:05
  • Actually, it's possible to take $n_1>2$ and $n_2>2$, since $n$ is neither a prime power nor twice a prime power. – Sávio Apr 08 '20 at 00:39
  • There is a problem here with powers of two, since $s$ may not be $\pm1 \pmod {2^k}$. For example, $(n,s)=(24,5)$ (so $5^2 \equiv 1 \pmod {24}$). We have $(n_1,n_2)=(6,4)$, but $\gcd(6,4)>1$. – Sávio May 22 '20 at 23:24