Let $1=b_1

Question

Let $1=b_1<b_2<\cdots<b_{\phi(n)}<n$ be integers relatively prime with $n$. How do I show that $$B_n =b_1\cdots b_{\phi(n)}\equiv\pm1\bmod n.$$

Answer 1

Verify the case $n=2$ directly and from now on assume $n>2$.

The set $\{b_1,\ldots,b_n\}$ forms an abelian group under multiplication modulo $n$, hence can be partitioned into pairs $(x,y)$ with $xy\equiv 1\pmod n$; these pairs effectively contribute irrelevant factors of $1$ to the product. The only numbers that cannot be paired are the numbers that are their own inverses, i.e. $x^2\equiv 1\pmod n$. These can be paired differently, namely $(x,y)$ with $x+y=n$ (note that $x^2\equiv 1$ implies $(n-x)^2\equiv 1$); now each pair contributes a factor of $xy\equiv -x^2\equiv -1$. This time, no number is left without partner because $x+y=n$ and $x=y$ implies $n=2x$, hence $\gcd(n,x)=\frac n2>1$, contradicting $\gcd(n,x)=1$.

Hence $b_1\cdots b_{\phi(n)}$ is congruent modulo $n$ to a product of factors $\pm1$, hence is so itself.

Answer 2

Take $\mathbb{Z}/\mathbb{nZ}^*$

Let $x\in \mathbb{Z}/\mathbb{nZ}^*$ ,$\exists $ an unique $y_x\in \mathbb{Z}/\mathbb{nZ}^*$ such that $xy_x=1$

Note that $x\to y_x$ is an isomorphism. So we have,

$\displaystyle(\prod _{x\in \mathbb{Z}/\mathbb{nZ}^*}x)(\prod _{x\in \mathbb{Z}/\mathbb{nZ}^*}y_x)=(\prod _{x\in \mathbb{Z}/\mathbb{nZ}^*}x)(\prod _{x\in \mathbb{Z}/\mathbb{nZ}^*}x)=1$

Let $y=\displaystyle (\prod _{x\in \mathbb{Z}/\mathbb{nZ}^*}x)$

Then we have $y^2=1\Rightarrow (y-1)(y+1)=0$ but as $y\in \mathbb{Z}/\mathbb{nZ}^*$ a field. So $y=\pm 1$

From this it follows that $\displaystyle (\prod _{x\in \mathbb{Z}/\mathbb{nZ}^*}x)=\pm 1$

From this the conclusion in the problem follows