prime divisor of $3n+2$ proof

Hint $ $ If $\,N = 3n+2\,$ has no prime factors of form $\,3k+2,\,$ then all prime factors must have form $\,3k+1,\,$ hence their product $\,N\,$ has form $\,3k+1,\,$ contra $\,N = 3n+2.$

Generally any factorization of an integer $\not\equiv 1\!\pmod{\!m}\,$ must include a factor $\not\equiv 1\!\pmod{\!m}\,$ simply because integers $ \equiv 1\!\pmod{\!m}\,$ are closed under multiplication (form a monoid). Similarly for other sets of integers closed under multiplication. Consider the set T of integers divisible by $10,\,$ i.e. those with unit digit $= 0.\,$ Any factorization of an integer $\,n\not\in T\,$ must include a factor $\not\in T,\,$ else all factors have unit digit $= 0\,$ thus so too does their product $\,n,\,$ i.e. $\,n\in T,\,$ contra hypothesis. Similarly, one could let $\,T\,$ be all powers of $\,10,\,$ etc. This property of factors of elements in the complement of $T$ is just an equivalent complementary restatement of the closure of $T$ under multiplication. See also this post on the complementary view of subgroups, which generalizes: integer $\times$ noninteger = noninteger; $\ $ rational $\times$ irrational = irrational, etc.

Hint:

Denote the prime factorization of $3n + 2$ by $$3n + 2 = p_1^{e_1}\cdot\ldots\cdot p_r^{e_r}$$ and consider this equation modulo $3$.

EDIT: Since other people have posted a full solution in the meanwhile, I'll do the same:

Modulo $3$ you get $$2 \equiv p_1^{e_1} \cdot\ldots\cdot p_r^{e_r}\pmod{3}.$$ Now assume that all prime factors $p_i$ are equivalent $0$ or $1$ modulo $3$. Then also $p_1^{e_1}\cdot\ldots\cdot p_r^{e_r}$ is equivalent to $0$ or $1$ modulo $3$. Contradiction.

We will assume that by number we mean positive integer.

Call a number of the form $3n+2$ which has no prime divisor of the form $3m+2$ bad. We will show that there are no bad numbers.

Suppose to the contrary that there is at least one bad number. Then there is a smallest bad number $b$.

It is clear that $b\ne 1$. It is also clear that $b$ is not prime. For if $b$ is prime, then it has a prime divisor of the form $3m+2$, namely itself. So there are integers $k$ and $l$, both bigger than $1$, such that $b=kl$.

If one of $k$ or $l$ is divisible by $3$, then so is their product. But $b$ cannot be divisible by $3$. For if $3$ divides $3n+2$, then $3$ must divide $2$, which is not the case.

If both $k$ and $l$ are congruent to $1$ modulo $3$, then so is their product. But $b$ is congruent to $2$ modulo $3$.

Thus at least one of $k$ and $l$ (in fact exactly one) is congruent to $2$ modulo $3$. Let it be $k$.

Because $k\lt b$, by the definition of $b$ as the smallest bad number, $k$ cannot be bad. It follows that $k$ is divisible by a prime $p$ of the form $3m+2$. But then so is $b$, contradicting the fact that $b$ is bad.

Remark: A proof that uses the fact that every integer $\gt 1$ is a product of primes is more efficient. Note that we do not need the Fundamental Theorem of Arithmetic, aka the Unique Factorization Theorem. It is enough to have a factorization.