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.

one of $n-1,n$ or $n+1$ is divisible by $3$. 
If it is $n$ then so is $n^2$.
If it is not $n$, then one of $n-1$ or $n+1$ is divisible by $3$, and hence so is their product $n^2-1$.
Thus, either $n^2$ or $n^2-1$ is a multiple of $3$. If$n^2+1$ would be a multiple of three, then one of $2=(n^2+1)-(n^2-1)$ or $1=(n^2+1)-n^2$ would be a multiple of three.
