Exercise 2.1.48* (Product of complete residue systems modulo $p$)

Proof. Let $\{r_1,r_2,\dots ,r_p\}$ and $\{r_1^{\prime },r_2^{\prime },\dots ,r_p^{\prime }\}$ be two complete residue systems modulo a prime $p>2$. Assume for contradiction that $\{r_1{r}_1^{\prime },r_2{r}_2^{\prime },\dots ,r_p{r}_p^{\prime }\}$ is a complete residue system modulo $p$.

As for every complete residue system, there exists one and only one index $i_0$ such that $r_{i_0}{r}_{i_0}^{\prime }\equiv 0(\mathit{mod}p)$. So $r_i{r}_i^{\prime }\not\equiv 0(\mathit{mod}p)$ if $i\ne i_0$, thus $r_i\not\equiv 0(\mathit{mod}p)$ and $r_i^{\prime }\not\equiv 0$ if $i\ne i_0$. Therefore $r_{i_0}\equiv 0\equiv r_{i_0}^{\prime }(\mathit{mod}p)$. Without loss of generality, we may assume that $i_0=p$, so that

are three reduced residue system modulo $p$.

By Exercise 47,

Then

Thus $1\equiv -1(\mathit{mod}p)$, so $p=2$. This is a contradiction since $p>2$ by hypothesis.

So $\{r_1{r}_1^{\prime },r_2{r}_2^{\prime },\dots ,r_p{r}_p^{\prime }\}$ cannot be a complete residue system modulo $p$. □