<s>[INST] 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.
 [/INST] 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 $a+2b \equiv 0 \pmod{3} \Rightarrow 3 \mid a+2b$.
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