Exercise 2.1.33 ($1^2,2^2,\ldots,m^2$ is not a complete residue system modulo $m$)

Question

Show that $1^2,2^2,\ldots,m^2$ is not a complete residue system modulo $m$ if $m>2$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
Note that $1^2 \equiv (m-1)^2 \pmod m$, and $1 
e m-1$ if $m>2$, so two distinct elements of the set $\{1^2,2^2,\ldots,m^2\}$ are congruent modulo $m$. Therefore this set is not a complete residue system modulo $m$ if $m>2$.
\end{proof}

\bigskip
Note: In other words, 
$$
\varphi
\left\{
\begin{array}{ccl}
\Z/m\Z& 	o & \Z/m\Z\
x & \mapsto & x^2
\end{array}
ight.
$$
is not bijective, since $\varphi([1]_m) = \varphi([-1]_m)$, where $[1]_m 
e [-1]_m$ if $m>2$.

Exercise 2.1.33 ($1^2,2^2,\ldots,m^2$ is not a complete residue system modulo $m$)

Answers

Comments

Exercise 2.1.33 ($1^2,2^2,\ldots,m^2$ is not a complete residue system modulo $m$)

Answers

Comments

Add answer