prove product of a reduced residue system is congruent to 1 mod $p^{\alpha}$

Question

Let $n=p^{\alpha}, p$ be an odd prime, $\alpha\in\mathbb{N}^*$. Assume that $\{x_1,...,x_k\}$ is a reduced residue system modulo $n$. Prove that $$x_1x_2...x_k\equiv -1\pmod{n}.$$

My attempt:
I have already shown that $x_1^2x_2^2...x_k^2\equiv 1\pmod{n}.$ So we have 2 cases here: $$x_1x_2...x_k\equiv -1\pmod{n}$$ or $$x_1x_2...x_k\equiv 1\pmod{n}.$$ I'm stuck at the part proving $x_1x_2...x_k\equiv 1\pmod{n}$ is impossible.
Could someone help me or have other ways to deal with the problem? Thanks in advance!

You can calculate exact formula for euler function $ \phi(n)$ — Adam Wrzesiński, Mar 22 '24 at 06:05
Does this answer your question? If ${a_1,\dots,a_{\phi(n)}}$ is a reduced residue system, what is $a_1\cdots a_{\phi(n)}$ congruent to? - found using an Approach0 search. Note that $N$, the # of solutions to $x^2\equiv 1\pmod{n}$, is $2$ in your ... — John Omielan, Mar 22 '24 at 06:11
(cont.) case, giving the RHS is $(-1)^{N/2}=-1$. FYI, Product of coprimes less than n shows the RHS is $-1$ if $n$ has a primitive root (which includes your $n=p^{\alpha}$ case), else it's $1$. In addition, the search also comes up with several other at least very closely related questions, such as The product of integers relatively prime to $n$ congruent to $\pm 1 \pmod n$. — John Omielan, Mar 22 '24 at 06:16

prove product of a reduced residue system is congruent to 1 mod $p^{\alpha}$

0 Answers0