Product of $x$ with $x$ and $x+1$ relatively prime to $n$

Question

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

Looking at $n=9$, $S=\{7,4,\}$ which I see are inverses modulo $9$. If I can prove that $x\in S \iff x^{-1} \in S$ we're done, but I'm unsure how to proceed with proving that?

Presumably, with $1\leq x\leq n$? – Thomas Andrews Jul 28 '17 at 00:43 — Thomas Andrews, Jul 28 '17 at 00:43

score 0 · Accepted Answer · answered Jul 28 '17 at 00:54

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Am I missing something here? If $x\in S$, then $xy\equiv 1 \bmod n$ for some $y$. Then $y+1=\frac{1}{x}+1=\frac{x+1}{x}$, a unit modulo $n$, since both $x, x+1$ are. Doesn't this prove what you need?

answered Jul 28 '17 at 00:54

Mohan

17,980

You do have to consider the possibility that $y = x$. – Robert Israel Jul 28 '17 at 01:06
@RobertIsrael If $y=x$, clearly $x^{-1}=x\in S$? – Mohan Jul 28 '17 at 01:27
But it won't appear twice so it can't cancel, so you need to argue that if $y=x$ then either $x \notin S$ or maybe $x \equiv 1.$ – CR Drost Jul 28 '17 at 01:29
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Right, if $y=x$ then $x^2=1$ and $(x-1)(x+1)\equiv 0\pmod n$. So $x\equiv 1\pmod{n}$ since is $x+1$ is invertible modulo $n$. – Thomas Andrews Jul 28 '17 at 01:32
Can you explain what is meant by "unit modulo $n$?" – meiji163 Jul 28 '17 at 20:57
What exactly do you mean by $x^{-1}\in S$ in your question, unless you mean modulo $n$? In real terms $x^{-1}$ is not even an integer. – Mohan Jul 28 '17 at 21:12
@Mohan I mean the multiplicative inverse, mod $n$: $x^{-1}x \equiv 1 \pmod n$ – meiji163 Jul 28 '17 at 23:01
Then units modulo $n$ makes sense. These are integers which are relatively prime to $n$, giving an invertible element modulo $n$. – Mohan Jul 29 '17 at 00:43
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@mysatellite I posted a complete proof here, in a recent question on this answer. – Bill Dubuque Dec 20 '20 at 15:28

Product of $x$ with $x$ and $x+1$ relatively prime to $n$

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