I found this generalization of the "$3 \mod 4" version while teaching number theory a few years ago.

Let $G$ be a proper subgroup of $(\mathbb{Z}/n)^\times$.  Then there are infinitely many primes $p$ such that $[p]\not\in G$.

*Proof:*  Suppose as usual that there are finitely many, $p_1, p_2, \ldots, p_r$, and find a number $g$ such that $(p_i,g) = 1$ for all $i$ and $[g]\not\in G$.  Then the number $N = np_1 p_2 \cdots p_r + g$ has a prime factorization $N = q_1q_2 \cdots q_s$ satisfying 

 - $q_i \neq p_j$ for all $i$ and $j$ and   
 - since $[N]=[g]\not\in G$, $[q_i]\not\in G$ for at least on $i$.