I guess this might as well be a separate answer. You can use quadratic reciprocity to give elementary proofs of certain special cases of Dirichlet's theorem. First, you should be aware of the following nice result and its "Euclidean" proof.

**Lemma:** Let $f(x) \in \mathbb{Z}[x]$ and let $P_f$ be the set of primes $p$ such that $p | f(n)$ for some $n$. Then $P_f$ is infinite.

*Proof.* If $f(0) = 0$ then this is obvious, so suppose otherwise. Let $p_1, p_2, ... p_n$ be a finite set of primes in $P_f$. Then for any $k$, $\frac{1}{f(0)} f(k f(0) p_1 ... p_n)$ must be divisible by a prime which is not one of the $p_i$, and choosing $k$ sufficiently large we can find a new prime $p_{n+1}$ in $P_f$.

An alternate proof is given here. Using this lemma and properties of the cyclotomic polynomials, you can prove that there exist infinitely many primes congruent to $1 \bmod n$ for any $n$ without any heavy machinery, so I will skip these cases.

Using quadratic reciprocity, you can prove that the following arithmetic progressions also contain infinitely many primes:

$11 \bmod 12$: Let $f(x) = 3x^2 - 1$. Then $p | f(n)$ implies $\left( \frac{3}{p} \right) = 1$. However, we can modify the proof of the lemma to prove that infinitely many of the primes dividing $f$ must be congruent to $3 \bmod 4$. To see this, let $p_1, .. p_n$ be finitely many primes with this property and consider $f(2 p_1 ... p_n) \equiv 3 \bmod 4$. It follows that infinitely many primes $p$ satisfy $\left( \frac{3}{p} \right) = 1$ and $p \equiv 3 \bmod 4$, so by quadratic reciprocity $\left( \frac{p}{3} \right) = -1$, so $p \equiv 2 \bmod 3$. Hence $p \equiv 11 \bmod 12$. In particular, there are infinitely many primes congruent to $2 \bmod 3$ and infinitely many primes congruent to $3 \bmod 4$.

$4 \bmod 5$: Let $f(x) = x^2 - 5$. Then $p | f(n)$ implies $\left( \frac{5}{p} \right) = 1$. Again, we can modify the proof of the lemma to prove that infinitely many of these primes are not congruent to $1 \bmod 5$. To see this, let $p_1, ... p_n$ be finitely many primes with this property, none of which are equal to $5$, and consider either $f(p_1 ... p_n)$ or $f(2 p_1 ... p_n)$, one of which is not congruent to $1 \bmod 5$ and which therefore has a prime factor which is not congruent to $1 \bmod 5$. So infinitely many primes $p$ satisfy $\left( \frac{5}{p} \right) = 1$ and $p \not \equiv 1 \bmod 5$. By quadratic reciprocity this gives $\left( \frac{p}{5} \right) = 1$, hence $p \equiv 4 \bmod 5$.

$3 \bmod 8$: Let $f(x) = x^2 + 2$. Then $p | f(n)$ implies $\left( \frac{-2}{p} \right) = 1$. Again, we can modify the proof of the lemma to prove that infinitely many of these primes are not congruent to $1 \bmod 8$. To see this, let $p_1, ... p_n$ be finitely many primes with this property, all of which are odd, and consider $f(2p_1 ... p_n) \equiv 6 \bmod 8$. So infinitely many primes $p$ satisfy $\left( \frac{-2}{p} \right) = 1$ and $p \not \equiv 1 \bmod 8$. By quadratic reciprocity this gives $p \equiv 3 \bmod 8$.

And so on. For what progressions is it possible to give these kinds of proofs? It turns out this is possible for the progression $a \bmod n$ if and only if $a^2 \equiv 1 \bmod n$. For details, see Keith Conrad's Euclidean proofs of Dirichlet's theorem.

More generally, quadratic reciprocity is the key to writing down the Dedekind zeta functions of quadratic number fields explicitly, and trying to generalize this leads you into class field theory and so forth.