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.

The key is that taking remainder modulo any fix number (now $3$) doesn't affect the operations $+,-,\cdot$, in that, for instance, if $a$ and $b$ gives the same remainder modulo $m$ (that is, $m$ divides $b-a$), then $a+c$ and $b+c$, or, $ac$ and $bc$ will also give the same remainder modulo $m$. The latter would also imply that $a^2$ gives the same remainder as $b^2$.
So, modulo $3$ now, as the remainders can only be $0,1,2$, if $x$ gives remainder $1$, then so does $x^2$. All we have to compute is $2^2=4$ gives again $1$ remainder. And of course, $0$ gives $0$.
 So, $2$ is not in the range of squaring modulo $3$.

The relation to give the same remainder modulo $m$ (that is, $m\,|\,b-a$) is called congruence and is denoted by
$$a\equiv b\pmod m$$
and also makes sense for negative numbers. For example, $-1\equiv 2\pmod{3}$, and that makes computation even easier: $(-1)^2=1$ (which confirms $-1\equiv 2\ \implies\ (-1)^2\equiv 2^2 \pmod3$ and that no $n^2$ is congruent to $-1$).
