TITLE: What's the explanation for why n^2+1 is never divisible by 3?
QUESTION [6 upvotes]: What's the explanation for why $n^2+1$ is never divisible by $3$?
There are proofs on this site, but they are either wrong or overcomplicated.
It can be proved very easily by imagining 3 consecutive numbers, $n-1$, $n$, and $n+1$. We know that exactly one of these numbers must be divisible by 3. 
$$(n-1)(n)(n+1)=(n)(n-1)(n+1)=(n)(n^2-1)$$
Since one of those first numbers had to have been divisible by $3$, this new product $(n)(n^2-1)$ must also be divisible by $3$. That means that either $n$ (and by extension $n^2$) or $n^2-1$ is divisible by $3$. If one of those has to be divisible by $3$, then $n^2+1$ cannot be. 
So it is definitely true. My question is why is this true, what is inherent about $1$ more than a square number that makes it not divisible by $3$? Another way of saying this might be to explain it to me as if I don't know algebra.

REPLY [9 votes]: Imagine a big square of dots. It has rows and columns. There is the same number of rows as columns. Plus there's one more dot.
Take three of the rows, and remove them. You have just removed a multiple of three dots - three from each column.  Now do the same thing for the rows. Now remove three columns. Again, you've removed a multiple of three. The number of rows and columns is the same again. There's still one more dot.
Keep doing this until there are just $0$,$1$, or $2$ rows and the same number of columns. All the dots you've removed so far can be grouped into a bunch of groups of three. So the square is either $0 \times 0$, $1\times 1$, or $2 \times 2$. So there are either $0+1= 1$, $1+1=2$, or $1+4=5$ dots left. None of these is a multiple of three. 
So the whole set of dots consists of a bunch of groups of three, and one group that's not a multiple of three. Clearly, it cannot be a multiple of three.