Find a modulus $m$ and a finite list of congruence classes $a\_1, ..., a\_r \mod m$ such that $(\frac{7}{n}) = 1 \iff n \equiv a\_i \mod m$ for some $i = 1$ to $r$.

Let me know if I am on the right track or give me some hints please.

Here is what I have so far: $\rightarrow$

Case 1: $n \equiv 1 \mod 4 \implies (\frac{7}{n}) = (\frac{n}{7}) = 1$

Thus the values for $n$ which $n$ is a quadratic residue $\mod 7$ are: $\{1, 2, 4\}$

Case 2: $n \equiv 3 \mod 4 \implies (\frac{7}{n}) = -(\frac{n}{7}) = 1$

Hence the values for which $n$ is a non-quadratic residue $\mod 7$ are: $\{3, 5, 6\}$

**EDIT**

case 1: $n \equiv 1 \mod 4$ and $n \equiv 4 \mod 7$ Now by the CRT we should be able able to find $n$.

Do: $n = 7 \cdot -1 \cdot 1 + 4\cdot 4 \cdot 2 = 25$. Checking we see $25 \equiv 1 \mod 4$ and $25 \equiv 4 \mod 7$ but I want $n \equiv$ either $1 \mod 28$ or $4 \mod 28$

So, what went wrong?