prime divisor of $3n+2$ proof 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.
 A: 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.
A: 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.
A: 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. 
A: 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.
A: I would try a proof by induction. The case $n=1$ yields $3n+2 = 5$, which is a prime so all is well. Now assume $n > 1$. If $3n+2$ is prime, we are again done. If not you can write it as $3n+2 = rs$ with $r,s < 3n + 2$. It is easy to show that either $r$ or $s$ also has to be of the form $3m + 2$. Applying the induction hypothesis gives the result.
A: OK, starting from the case 2:
$$3n+2=r(3m+1)$$
$$3(n-mr)=r-2$$
$$k=n-mr=\frac{r-2}{3}$$
$$r = 3k+2$$
Is this correct?
