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?