Number theory: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system Is this an acceptable answer?
Question:  show that  ${ 1^2, 2^2, 3^2,... , m^2}$  cannot be a complete residue system.
Since the above has $m$ elements, one must show it cannot be a complete residue system
modulo $m$. Consider the complete residue system modulo m:
{1,2,3,... ,$m$}. Now between the two the first element $1^2$ and $1$ matches. But
what about the second last element $ (m-1)^2$. Here we have $(m-1)^2$ = $m^2 -2m +1$
which is clearly congruent to the first element, $1$, modulo m.
So there exist two elements congruent modulo $m$ to each other in ${ 1^2, 2^2, 3^2,... , m^2}$ so it cannot be a complete residue system. 
 A: Yes, it is correct: because you have found two congruent elements in the set $\{1^2,\ldots,m^2\}$, it cannot form a complete residue system (which would mean that we have $m$ pairwise incongruent numbers).
However when $m\leq2$, then $m-1\equiv1\pmod m$ and you don't get a contradiction from showing $(m-1)^2\equiv1^2$.
So it's correct for $m>2$.
A: I think $1^{2},2^{2},....,m^{2}$ is not a complete residue system when $m$ is an odd.
For this it sufficent to prove that two elements of this set are congruent modulo each other
A: Let's start with an example. Below is a congruence table modulo &11&.
\begin{array}{rrr}
 k & k^2 & k^2 \pmod{11} \\
\hline
 0 &    0 &   0\\
 1 &    1 &   1\\
 2 &    4 &   4\\
 3 &    9 &   9\\
 4 &   16 &   5\\
 5 &   25 &   3\\
 6 &   36 &   3\\
 7 &   49 &   5\\
 8 &   64 &   9\\
 9 &   81 &   4\\
10 &  100 &   1\\
\hline
\end{array}
Notice that the last column does not contain all eleven of the integers from $0$ to $10$ because, of the eleven numbers, there are repeated numbers: $k^2 \equiv (11-k)^2 \pmod{11}$.
This is true for all moduli, N.
$$(N-k)^2 \equiv N^2 - 2kN + k^2 \equiv k^2 \pmod N$$
