TITLE: prime divisor of $3n+2$ proof
QUESTION [14 upvotes]: I have to prove that any number of the form $3n+2$ has a prime factor of the form $3m+2$. Ive started the proof
I tried saying by the division algorithm the prime factor is either the form 3m,3m+1,3m+2.
Case 1: $3m$
     suppose $3m$ does divide $3n+2$. That means $3mr=3n+2$ where $r$ is some integer. But we get $mr-n= 2/3$ and since $mr-n$ is an integer that's a contradiction. Thus $3m$ does not work.
Case 2: $3m+1$
Same argument here
Case 3: $3m+2$
By the fundamental theorem of arithmetic every integer is divisible by a prime. But not sure what to do from here.

REPLY [3 votes]: If $(3n+2)$ is prime, then it has a prime factor $(3n+2)$.
If $(3n+2)$ is composite, then there exist integers $h,k$ where $1< h,k <(3n+2)$.
Case 1: if at least one of $h,k$ be of the form $3m+2$ then there is nothing to prove.
Case 2: if at least one of $h,k$ be of the form $3m$ then, $3|(3n+2)$
i.e. $3|2$ a contradiction. So 
any integer of the form $(3n+2)$ has no prime factor of the form $3m$.
case 3:if at least one of $h,k$ be of the form $3m+1$ (say $h$).
then $3$ cases arises for $k$
subcase 1: $k= 3p$ which is impossible by case $2$.
subcase 2: $k=3j+1$, then $3n+2 = 3(3mj+m+j)+1$ which is again a contradiction.
subcase 3: so we must have $k$ of the form $3m+2$.
combining all the cases we have the required result.