Product of $x$ with $x$ and $x+1$ being coprime to $n$

Question

For an odd positive integer $n \gt 1$, let $S$ be the set of integers $x$, $1 \le x \le n$, such that both $x$ and $x+1$ are relatively prime to $n$. Show that $$\prod_{x\in S} x \equiv 1 \pmod n$$

I know there is already a solution on this site here but I don't understand the solution fully and the thread was created over 3 years ago so that is why I decided to create a new thread. Can anyone retranslate the solution presented in the linked thread. On my approach I thought of using $CRT$ but I couldn' t solve the problem using it, so if you know solution that involves $CRT$ please post it.

Any help appreciated.

Answer 1

Lemma $ $ If $\,S\,$ is a subset of $\,\Bbb Z_n\,$ closed under inverses then the product of its elements equals the product of the set of roots of $\,x^2-1\,$ lying in $\,S$.

Proof $\ S$ partitions into pairs $\,(x,x^{-1})$ with $\,x\not\equiv x^{-1}$ & singletons $(x)$ with $\,x^{-1}\equiv x.\,$ Pairs contribute $\,1\,$ to the product, so it reduces to the product of the singletons = roots of $\,x^2\equiv 1$.

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Your $S\,$ is closed under inverses, i.e. $\,(n,x(x\!+\!1))=1\Rightarrow\,(n,x^{-1}(x^{-1}\!+\!1))=1,\,$ since $\,x^2\, x^{-1}(x^{-1}\!+\!1) = 1+x\,$ is coprime to $\,n\,$ hence so too are its factors $\,x^{-1},\ x^{-1}\!+\!1.$

If $\,x\in S\:\!$ and $\,x^2\!-1\equiv (x\!-\!1)(x\!+\!1)\equiv 0\, $ then $\,x\!-1\equiv 0,\,$ by $\,x\!+\!1\,$ is cancellable, being coprime to $\,n.\,$ So the set of roots of $\,x^2\!-1\,$ in $\,S\,$ is $\{1\},\,$ with product $1 = \prod S\,$ via Lemma.

Remark $ $ This is closely related to this elementary proof of a generalization of Wilson's theorerm to $\,\Bbb Z_n,\,$ based on similar reflection (involution) symmetry afforeded by inversion.

Answer 2

Since $x$ is relatively prime with $n$ there exists some $y$ s.t. $xy \equiv 1 \mod {n}$. Now assume that $x \ne y \mod{n}$. Then observe that $y+1 \equiv \frac {1}{x} +1 \equiv \frac {1+x}{x}\mod {n}$. Since both $x+1$ and $x$ are relatively prime with $n$, they have muliplicative inverses respective to modulo $n$. Thus, from the previous equality, $y+1$ necessarily has a multiplicative inverse. But this implies that $ y+1 $ and $n$ are relatively prime. Moreover $ y $ and $n$ are relatively prime(since $x$ is the multiplicative inverse of $y$).Hence $y \in S$. If $x = y$, then $x^2 \equiv 1 \mod{n}$ which implies $(x-1)(x+1) \equiv 0 \mod{n}$. But since $x+1$ is relatively prime with $n$, it has an inverse and so this implies that $x\equiv 1 \mod {n}$. Hence, if $x \in S $ and x is not equal to its multiplicative inverse $y$ in modulo $n$, $y \in S$,else $x\equiv 1 \mod {n}$ .