How to prove the statement? 
Say if is false or true that:

The sum or subtraction of two positive integer numbers not divisible by $3$ is always a
    multiple of $3$.


My attempt was:
I think that is true, because i can't get counterexamples for this statement, so i'll try to prove it.
Let $a,b \in \mathbb{Z^+}$ the two numbers. 
And, $\frac{a}{3} = \frac{p}{q}$, same for $b$,  $\frac{b}{3} = \frac{m}{n}$
Where $p,m \in \mathbb{Z}^+ - \{0\}$ and $q,n \in \mathbb{Z}^+ -\{0,1\}$  and $p\neq q, m\neq n$, Also $\frac{p}{q}, \frac{m}{n}$ are simplified. These conditions ensure that $\frac{p}{q}, \frac{m}{n}$ are real numbers and not integers.
The sum of these numbers is equal to $3(\underbrace{\frac{p}{q}}_\text{a} +\underbrace{\frac{m}{n}}_\text{b})$
The substraction of these numbers is equal to  $3|(\frac{p}{q} - \frac{m}{n})|$
Thus i need to prove that the sum $\frac{pn+mq}{qn}$ is integer or the substraction $|(\frac{pn-mq}{qm})|$ is a integer.
But here I have arrived. Any hint is appreciated.
 A: Let $a, b$ be two integers not divisible by $3$. Then, each of $a$ and $b$ are either $1$ or $2$ mod $3$.
If they are the same mod $3$, then $a-b\mod3=0$ so $a-b$ is divisible by 3.
If they are different mod $3$, then $a+b\mod3=0$ so $a+b$ is divisible by 3.
A: True
We can represent these integers, not divisible by 3, as $3k+1$ and $3t+2$, $k,t \in \mathbb{Z}$. Now, look at the 4 possible $\pm$ cases. For simplicity, only the divisible case is written  


*

*$3k+1 \pm 3t +1$


*

*$3k+1 - 3t -1 = 3k -3t$ and $3| 3(k-t)$


*$3k+2 \pm 3t +2$


*

*$3k+2 - 3t -2 = 3k -3t$ and $3| 3(k-t)$


*$3k+1 \pm 3t +2$


*

*$3k+1 + 3t +2 = 3(k+t) +3$ and $3 | 3(k+t) +3$


*$3k+2 \pm 3t +1$


*

*$3k+2 + 3t +1 = 3(k+t) +3$ and $3 | 3(k+t) +3$
For each case, we have a either plus or minus is divisible. Therefore the statement is true.
A: $\bmod 3\!:\ a,b\not\equiv 0\,\Rightarrow\, a,b\equiv \color{#c00}{\pm1}\,\Rightarrow\,0\equiv\overbrace{ a^2-b^2}^{\textstyle\color{#c00}{ 1- 1}}\equiv (a\!-\!b)(a\!+\!b)$ $\overset{3\ \rm prime}\Longrightarrow \color{#0a0}{a\!-\!b}\equiv 0\,$ or $\,\color{#90f}{a\!+\!b}\equiv 0$
Remark $\ $ More generally applying little Fermat similarly yields
$\bmod p\!:\ a,b,\color{#0a0}{a\!-\!b}\not\equiv 0\,\overset{\rm Fermat}\Longrightarrow\,0\equiv\overbrace{ a^{p-1}-b^{p-1}}^{\textstyle\color{#c00}{ 1- 1}}$ $ \equiv (\color{#0a0}{a\!-\!b})\dfrac{a^{p-1}-b^{p-1}}{a-b}$ $\overset{p\ \rm prime}\Longrightarrow\, \color{#90f}{\dfrac{a^{p-1}-b^{p-1}}{a-b}}\equiv 0$
