Is my proof consider to be correct? Problem: Let $a,b ∈ \mathbb N$, prove that at least one of the following $ab, a+b, a-b$ is evenly divisible by 3.
My solution:
Case 1: If $a_{mod}3=0 \text{ or } b_{mod}3=0$ then we can say that $a = 3k,\ ab=3kb$ which is evenly divisible by three
Case 2: If $a_{mod}3=b_{mod}3$ then $a=3k_1+a_{mod}3,\\b=3k_2+b_{mod}3,\\a-b=3k_1+a_{mod}-(3k_2+b_{mod}3)\\a-b=3k_1+a_{mod} -3k_2-b_{mod}3\\\text{since }a_{mod}3=b_{mod}3\\a-b=3k_1-3k_2\\a-b=3(k_1-k_2)$ 
in this case we are prove that $a-b$ is evenly divisible by three.
Case 3: If $a_{mod}3$ doesn't equal to $b_{mod}3$, if this stands for true, there are two combinations 
$a_{mod}3=1, b_{mod}3=2 \text{ and}\\a_{mod}3=2, b_{mod}3=1\\
\text{if this stands for true we can prove that } a+b \text{ is divisible by three}\\
a+b=3k_1+a_{mod}3+3k_2+b_{mod}3, \text{since } a_{mod}3+b_{mod}3=3\\
a+b=3k_1+3k_2+3\\
a+b=3\cdot(k_1+k_2+1)$ we proved for all cases.
Is my proof correct?
 A: This proof looks correct to me, well done!
The one critique I would make is notational: $a_{mod}b$ is a notation that is extremely uncommon and one I would recommend avoiding. Bill Dubuque suggests that it represents the remainder left over when dividing $a$ by $b$ (I had assumed it meant the image of $a$ under the natural map $\mathbb{Z}\to\mathbb{Z}/b\mathbb{Z}$) which is certainly something that you should explicitly spell out. Generally I would  recommend you avoid using remainders and just use modular arithmetic, writing 3 = 0 \pmod{b} to produce $3=0\pmod{3}$ instead. An uncommon notation similar to the one you're employing would be $[3]_3=[0]_3$, though I would also discourage using this notation due to its rarity.
A: Seems correct, but if you're allowed to use Fermat's little theorem, you could go like this :
If $a$ or $b$ are divisible by $3$ then it is obvious that $ab$ is divisible by $3$.
If neither $a$ or $b$ are divisible by $3$, then let's take a look at $(a-b)$ and $(a+b)$. Take their product : $(a-b)(a+b) = a^2 - b^2$ and rewrite it as : $(a^2-1) - (b^2-1)$. 
By Fermat's LT, both $a^2-1$ and $b^2-1$ are divisible by $3$, so is their difference. This means that $(a-b)(a+b)$ is divisible by $3$. Therefore, either $a-b$ or $a+b$ is divisible by 3.
