<s>[INST] Why is $n^2+4$ never divisible by $3$? Can somebody please explain why $n^2+4$ is never divisible by $3$?  I know there is an example with $n^2+1$, however a $4$ can be broken down to $3+1$, and factor out a three, which would be divisible by $4$.
 [/INST] Well, You have two cases
1st case $n$ is odd then $n=2k+1$ for some integer $k$
Then $n^2 + 4 = (2k+1)^2 + 4 = 4k^2 + 4k + 5 = 4k(k+1) +5$
Here we have either $k$ or $k+1$ is even and the other is odd or vice versa and we know that even $\times$ odd = even and so $4k(k+1) + 5 = 4(2m)+5$ for some integer $m$ and so have $8m +5$ Now it's obvious that $3 \nmid 8m+5$ for any integer $m$
2nd case $n$ is even then $n=2k$ for some integer $k$ and so $n^2 +4 = (2k)^2 + 4 = 4k^2 +4 = 4(k^2 +1)$
You should also deduce that 3 doesn't divide $4(k^2 + 1)$ and you are basically done !
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