Let $a,b \in \mathbb Z$. Prove that if $3 \mid (a+2b)$ then $3 \mid (2a+b)$ 
Let $a,b \in \mathbb Z$. Prove that if $3 \mid (a+2b)$ then $3 \mid (2a+b)$

This is how I solved this: 
$3m = a+2b \iff a = 3m-2b$ 
$2a+b = 2(3m-2b)+b = 6m -3b = 3(2m-b)$
 And now, my solution seems to work. However, when - instead of solving for $a$, I solve for $b$, then I get this: 
$2a+b = 3 \frac{a+m}{2}$ 
How can I even be sure that this number is an integer? 
Is there a way to fix the second solution? If my first solution is correct. the second one should work as well.
 A: $3|a+2b$ then $3|2a+4b$ then $3|2a+4b-3b = 2a+b$.
Of course, you can also reverse this. If $3|2a+b$ then $3|2a+4b = 2(a+2b)$ Now use Euclid lemma and we have $3|a+2b$.
A: Since $a-3m=-2b$, $a+m = 4m-2b = 2(2m-b)$ which is even and thus is divisible by two. Both of your solutions work, but the first one is cleaner since it does not involve the division by two.
A: Let me actually address your question rather than just giving an answer.
Both solutions work. For the second one:

when - instead of solving for $a$, I solve for $b$, then I get this: 
  $2a+b = 3 \frac{a+m}{2}$ 
  How can I even be sure that this number is an integer?

To complete the argument, notice that $2a + b$ is an integer (obviously); therefore, $3 \frac{a + m}{2}$ is an integer. And because it's an integer, $\frac{a + m}{2}$ must be an integer (since $3$ does not cancel $2$). This means that it is a multiple of $3$.
A: It's easy to observe that 
$$3a+3b \equiv 0 \pmod{3} \Leftrightarrow a+2b \equiv -2a-b \pmod{3} \tag{1}$$
and the relation becomes $\iff$.
$\color{red}{\Rightarrow}$ If $3 \mid a+2b$ then $a+2b \equiv 0 \pmod{3}$ and from $(1)$ we hqve $0 \equiv -2a-b \pmod{3}$ or $0 \equiv 2a+b \pmod{3} \Rightarrow 3 \mid 2a+b$.
$\color{red}{\Leftarrow}$ If $3 \mid 2a+b$ then $2a+b \equiv 0 \pmod{3}$ and from $(1)$
$$-a-2b \equiv 2a+b \pmod{3} \Rightarrow -a-2b \equiv 0 \pmod{3}$$
or $a+2b \equiv 0 \pmod{3} \Rightarrow 3 \mid a+2b$.
A: Why don't you use congruences ? They usually show the right short way in this kind of divisibility questions. Here you want to prove that $a+2b \equiv 0$ mod $3$ iff $2a+b \equiv 0$ mod $3$. But $2\equiv -1$ mod $3$, so the two conditions are immediately equivalent to $a\equiv b$ mod $3$.
