There is no such set ever

Consider the sequence of squares $\displaystyle 0,1,4,9,16,25,36,49,64,81,100,...$ in modulo:

$\displaystyle \bmod 7: 0,1,4,2,2,4,1,0,1,4,2,2,4,1,0,1,...$ (notice a pattern?)

Notice that $\displaystyle (n+7)^2-n^2=14n+7 \equiv 0 (\bmod 7)$, so the set repeats in blocks of $\displaystyle 7$.

Also, $\displaystyle (7k+r)^2-(7k-r)^2=28kr \equiv 0 (\bmod 7)$ so the set "reflects" as shown in this example. More clearly, for any $\displaystyle k$, $\displaystyle (7k)^2 \equiv 0 (\bmod 7)$, therefore $\displaystyle (7k+r)^2 \equiv (7k-r)^2 (mod 7)$ for all $\displaystyle r$.

Both of these observations can be generalized. For any prime $\displaystyle p$,

$\displaystyle (n+p)^2-n^2 \equiv 0 (\bmod p)$

$\displaystyle (pk+r)^2-(pk-r)^2 \equiv 0 (\bmod p)$

For all appropriate $\displaystyle n,k,r$

Therefore, the size of the set of residues of $\displaystyle n^2$ for any prime $\displaystyle p$ is at most $\displaystyle \lceil \frac{p}{2} \rceil$

Sorry