Show that there exists infinitely many primes which satisfy a given congurence. Let $m$ be a fixed positive integer that is the product of distinct prime factors of the form $(3k+2)$, such as $5 \times 11$. 
Prove that there exist infinitely many primes $p$ such that $3^{3p-2}\equiv 1 \pmod m$?
I started with assuming that there exists finitely many primes satisfying such equation. How to contradict  this statement? May be we can create one more prime satisfying given equation? I am stucked. Please help
 A: Your requested statement to prove that there exists infinitely many primes $p$ such that
$$3^{3p - 2} \equiv 1 \pmod{m} \tag{1}\label{eq1A}$$
is not always true based just requiring $m$ being a product of distinct prime factors of the form $3k + 2$. One way to see this is that for odd primes $p$, where $p = 2j + 1$ for some $j$, that $3p - 2 = 3(2j + 1) - 2 = 6j + 1$, so \eqref{eq1A} becomes
$$3^{6j + 1} \equiv 1 \pmod{m} \tag{2}\label{eq2A}$$
Multiplying both sides by $3$ gives
$$\begin{equation}\begin{aligned}
3^{6j + 2} & \equiv 3 \pmod{m} \\
\left(3^{3j + 1}\right)^2 & \equiv 3 \pmod{m}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The above congruence relation being true for modulo $m$ means it must also be true for each factor of $m$, so this means each prime factor of $m$ must have $3$ being a quadratic residue. However, as shown in the table of the Law of quadratic reciprocity section, $3$ is a quadratic residue for an odd prime $q$ (so the only exception is for $2$ being one of the factors of $m$) if and only if $q \equiv 1, 11 \pmod{12}$. Only $q \equiv 11 \pmod{12}$ meets the requirement that the primes are of the form $3k + 2$, but it also requires $q \equiv 3 \pmod{4}$. Thus, if any of the odd prime factors of $m$ are of the form $q \equiv 1 \pmod{4}$ instead, then \eqref{eq3A}, and thus \eqref{eq1A}, will not be true.
For example, $5 \equiv 1 \pmod{4}$ meets the requirement of being a prime of the form $3k + 2$, so let $m = 5$. The left side of \eqref{eq2A} gives
$$3^{6j + 1} \equiv 3^{1}(3^{6})^{j} \equiv 3(-1)^{j} \equiv 2,3 \pmod{5} \tag{4}\label{eq4A}$$
In particular, it's never congruent to $1$.
