Exercise 3.8

Proof. $\text{∙}$ existence.

As $p$ is prime and $1\le k\le p-1$, $k\wedge p=1$, so there exist ${\lambda }_k,{\mu }_k\in \mathbb{Z}$ such that ${\lambda }_{k}p+{\mu }_{k}k=1$. Let $b_k\in \{0,1,\dots ,p-1\}$ such that $b_k\equiv {\mu }_k(\mathit{mod}p)$. Then $k{b}_k\equiv 1$, and $b_k\not\equiv 0(\mathit{mod}p)$, so $1\le b_k\le p-1$.

$\text{∙}$ unicity. If $k{b}_k\equiv k{b}_k^{\prime }(\mathit{mod}p)$, where $b_k,b_k^{\prime }\in \{1,2,\dots ,p-1\}$, then $p\mid k(b_k^{\prime }-b_k)$, and $p\wedge k=1$, thus $p\mid b_k^{\prime }-b_k$. $b_k^{\prime }\equiv b_k$, and $b_k,b_k^{\prime }\in \{1,2,\dots ,p-1\}$, so $b_k=b_k^{\prime }$.

If $p$ is a prime number, and $k\in \{1,2,\dots ,p-1\}$, there is a unique $b_k$ in $\{1,2,\dots ,p-1\}$ such that $k{b}_k\equiv 1(\mathit{mod}p)$.

If $k=b_k$, then $k^2\equiv 1(\mathit{mod}p)$, so $p\mid (k-1)(k+1)$, and $p$ is a prime, thus $p\mid k-1$ or $p\mid k+1$, that is $k\equiv \pm 1(\mathit{mod}p)$. As $1\le k\le p-1$, $k=1$ or $k=p-1$ (and $1^2\equiv {(p-1)}^2\equiv 1(\mathit{mod}p)$). □