Timeline for Is it possible to have square-free order(s) in $\mathbb{Z}^\times_N$?
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| May 11 at 14:32 | answer | added | Somudro Gupto | timeline score: 0 | |
| Apr 17 at 6:20 | comment | added | Somudro Gupto | @JohnOmielan Yes, you are right. Later, I noticed that both $\pm 1\pmod N$ are impossible as $p_i\ll p',q'$. The proof is here. First, for brevity, let's denote $w=p_0^{p_1\times p_2\times\cdots\times p_m}\pmod N$. Now, the only element that has order $1$ in $\mathbb{Z}^\times_N$ is the identity $1$ itself. Since, $p_0 \ne 1$, $w \neq 1$. For $w=-1$, we need either $p_0 \equiv -1 \pmod{p}$ or $p_0 \equiv -1 \pmod{q}$ or both. So, $p_0 \ge p-1=2p'$ or $p_0 \ge q-1=2q'$. It contradicts with $p_i\ll p',q'$. | |
| Apr 16 at 15:02 | comment | added | John Omielan | You are welcome. Actually, $m_1=1$ is also a possible order, with $m_1=2$ only being the order in the $p_0^{p_1\times p_2\times\cdots\times p_m}\equiv -1\pmod{N}$ case. Regardless, with $m_1=1$ or $m_1=2$, based on the problem conditions, consider what that means about the possibility of such odd primes $p_i$ existing. | |
| Apr 16 at 5:32 | comment | added | Somudro Gupto | @JohnOmielan Thank you for your answer. So the only possible order is $m=2$. | |
| Apr 15 at 19:22 | comment | added | John Omielan | Note that, in the comment above, my $m$ should be something different (e.g., $m_1$) to avoid conflicting with your $m$. | |
| Apr 15 at 18:41 | review | Close votes | |||
| Apr 17 at 16:42 | |||||
| Apr 15 at 18:18 | comment | added | John Omielan | Welcome to MathOverflow. With the Multiplicative order, letting $m=\operatorname{ord}_{N}(p_0)=m$, then note $m\mid 2p_1\times p_2\times\cdots\times p_m$ and $m\mid \varphi(N)$ where $\varphi(N)=(2p')(2q')=4p'q'$. | |
| Apr 15 at 15:50 | history | edited | Somudro Gupto | CC BY-SA 4.0 |
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| Apr 15 at 15:33 | history | edited | Somudro Gupto | CC BY-SA 4.0 |
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| S Apr 15 at 15:32 | review | First questions | |||
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| S Apr 15 at 15:32 | history | asked | Somudro Gupto | CC BY-SA 4.0 |