Quadratic reciprocity modulo $2$ works slightly differently. In fact, it holds that
$$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}.$$
Thus, you have:
$$\left(\frac{14}{p}\right) = (-1)^{(p^2-1)/8} \cdot (-1)^{(p-1)/2} \cdot \left(\frac{p}{7}\right).$$
This means that you need to look at the form of $p$ modulo $8$ (first two terms) and modulo $7$ (last term). For $p \equiv 1, 3, 5, 7 \pmod{8}$, the product of the initial two terms is $+1, +1, -1, -1$ respectively. Thus, you need either of the two options to hold:
$p \equiv 1,3 \pmod{8}$ and $p$ is a quadratic residue modulo $7$, i.e. $p \equiv 1,2,4 \pmod{7}$,
$p \equiv 5,7 \pmod{8}$ and $p$ is a not quadratic residue modulo $7$ i.e. $p \equiv 3,5,6 \pmod{7}$.
In each situation, you have $6$ possible cases, each corresponding to a "good" residue of $p$ modulo $56$ (by Chinese remainder theorem). It is a little mundane to work these out, but each pair of congruences $p \equiv a \pmod{8}, p \equiv b \pmod{7}$ is equivalent to the single congruence $p \equiv c \pmod{56}$, where $c$ happens to be given by $c = 8 b - 7 a$.
Given ODD primes $p \neq q:$ and $$ \color{magenta}{p \equiv 3 \pmod 4} $$
Lemma: $$ (-p|q) = (q|p). $$
Lemma: If $$ a^2 + p \equiv 0 \pmod q, $$ THEN $$ (q|p) = 1. $$
Let $$ F_1 = 4 + p, $$
$$ F_2 = 4 F_1^2 + p, $$
$$ F_3 = 4 F_1^2 F_2^2 + p, $$
$$ F_4 = 4 F_1^2 F_2^2 F_3^2 + p, $$
$$ F_5 = 4 F_1^2 F_2^2 F_3^2 F_4^2 + p, $$
and so on.
These are all of the form $a^2 + p$ and are odd, so the only primes than can be factors are quadratic residues for $p.$ Next, all the $F_j$ are prime to $p$ itself. Finally, these are all coprime. So, however they factor, we get an infinite list of primes that are quadratic residues of $p.$
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Given ODD primes $p \neq q:$ and $$ \color{magenta}{p \equiv 1 \pmod 4} $$
Lemma: $$ (p|q) = (q|p). $$
Lemma: If $$ a^2 - p \equiv 0 \pmod q, $$ THEN $$ (q|p) = 1. $$
FIND an even square $$ W = 4^k = \left( 2^k \right)^2 $$ such that
$$ \color{magenta}{ W > p.} $$
Let $$ F_1 = W - p, $$
$$ F_2 = W F_1^2 - p, $$
$$ F_3 = W F_1^2 F_2^2 - p, $$
$$ F_4 = W F_1^2 F_2^2 F_3^2 - p, $$
$$ F_5 = W F_1^2 F_2^2 F_3^2 F_4^2 - p, $$
and so on. As $p \equiv 1 \pmod 4$ and $W \equiv 0 \pmod 4,$ we know $W - p \equiv 3 \pmod 4 $ and so $W-p \geq 3. $ So the $F_j$ are larger than $1$ and strictly increasing.
These are all of the form $a^2 - p$ and are odd, so the only primes than can be factors are quadratic residues for $p.$ Next, all the $F_j$ are prime to $p$ itself. Finally, these are all coprime. So, however they factor, we get an infinite list of primes that are quadratic residues of $p.$
Best Answer
This is just a tiny bit more involved than Wolfram Alpha can handle:
Select[Prime[Range[100]], Union[JacobiSymbol[Prime[Range[12]], #]] == {-1} &]The answer is $163$. You could have gotten the answer a lot quicker if you had just asked a wood nymph.
Oh, right, you don't know how to talk to wood nymphs. Okay, so if in Mathematica you change
100to4096, the only answer is $163$. However, go up a little higher to find $74093$.Why so few? That might be a more interesting question. Or not.