Prove that $n^2 + 1$ is not a multiple of $6$ for any positive integer $n$

Question

Prove that $n^2+1$ is not a multiple of $6$ for any positive integer $n$. I i think prime factorization would be a good way to go about this problem but I need some help.

Answer 1

$$n^2+1\equiv0\pmod6$$

$$\implies n^2+1\equiv0\pmod3\iff n^2\equiv-1$$

But, $\displaystyle n\equiv0\pm1\pmod3\implies n^2\equiv0,1\not\equiv-1$

Answer 2

$-1$ is not a quadratic residue modulo $3$.

Answer 3

$n^2+1$ is not a multiple of $3$. To see this, consider three cases where $m$ is of the form $3k, 3k+1$ and $3k+2$ respectively. Squaring, you'll get the first one is of the form $9k^2\equiv3k^2$, the other two are $9k^2+6k+1=3(3k^2+2k)+1\equiv3k^2+1$ and $9k^2+12k+4=3(3k^2+4k)+1\equiv3k^2+4 \equiv3k^2+1$.

Thus, $n^2$ is of the form $3l$ or $3l+1$, implying $n^2+1$ will be of the form $3l+1$ or $3l+2$, none of which is a multiple of $3$.

Since $3$ is a factor of $6$, $n^2+1$ will never be a multiple of $6$ either.

$\equiv$ sign means they are equivalent when dividing by $3$.