What's the explanation for why n^2+1 is never divisible by 3? 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.

Another approach is to write $n=3k+r$, with $r=0$, or $r=1$, or $r=2$, which is always possible: it's a simple division with the remainder. Now, the case $r=0$ is easily settled:
$$
n^2+1=9k^2+1
$$
which can't be divisible by $3$.
If $r=1$, then
$$
n^2+1=(3k+1)^2+1=9k^2+6k+1+1=3(3k^2+2k)+2
$$
which is not divisible by $3$.
If $r=2$, then
$$
n^2+1=(3k+2)^2+1=9k^2+12k+4+1=3(3k^2+4k)+3+2=3(3k^2+4k+1)+2
$$
and again this is not divisible by $3$.
