How do I prove that numbers not divisible by 3 can be represented as 3x+1 or 3x-1? I saw that some proofs used the fact that numbers not divisible by $3$ can be represented as $3x+1$ or $3x-1$. But how do I prove that it is true?
 A: Remainder theorem:
Take $n$, divide by $3$. The remainder is either $0$ or $1$ or $2$.
If remainder is $1$, then $n=3q+1$ by long divison if you like.
If remainder is $2$ then $n=3x+2=3(x+1)-1$.
A: If $k=3x+1$ then $$k\equiv 1\mod 3$$ and if $k=3x-1$ then $$k\equiv 2\mod 3.$$ Moreover $$\mathbb Z/3\mathbb Z=\{[0],[1],[2]\}$$
where $$[1]=\{3x+1\mid x\in\mathbb Z\}$$ and $$[2]=\{3x+2\mid x\in\mathbb Z\}=\{3x-1\mid x\in\mathbb Z\}$$
what conclude the proof. 
A: Do you think it is true? Do you understand what the statement means and just have problems writing a formal proof, or is the statement itself meaningless to you? 
As an example, take x = 12. There is 3x = 36, 3x-1 = 35, 3x+1 = 37. So there is a range of three numbers which are either divisible by 3 or equal to 3x-1 or 3x+1. 
Now take x = 13. There is 3x = 39, 3x-1 = 38, 3x+1 = 40. Three numbers from 38 to 40. Just before we had three numbers from 35 to 37. So these ranges fit nicely together giving a range from 35 to 40 of six numbers which are either divisible by 3 or equal to 3x-1 or 3x+1. 
Now take x = 14, and the range increases to 35 to 43. Does this make sense now? The numbers divisible by 3 are three apart (..., -6, -3, 0, 3, 6, ... ). Each of the two numbers within the gap between two of these numbers is either one more than the previous or one less than the next number. 
Maybe you should now think about whether it is true or not that every number not divisible by 4 can be represented as 4x-1 or 4x+1. 
