$3$ never divides $n^2+1$

Solution 1:

Instead of considering whether $n$ is even or odd, consider the remainder when $n$ is divided by $3$; as an example, if the remainder is $1$, we have $$n = 3k + 1 \implies n^2 + 1 = 9k^2 + 6k + 2$$

which is not divisible by $3$. There are two more cases.

Solution 2:

Another way: $ $ notice that $\,3\,$ divides one of $\ \color{#0a0}{n\!-\!1,\,n,\,n\!+\!1}.\ $ Therefore

$\ \ \ \color{#c00}{3\mid n^2\!+1}\Rightarrow\ 3\mid 2 = (\color{#c00}{n^2\!+1})(2\!-\!n^2)+\color{#0a0}{(n\!-\!1)n^2(n\!+\!1)},\ $ contradiction.

Remark $\ $ The above implies coprime $\,n^2\!+1\,$ and $\,n^3\!-n = (n\!-\!1)n(n\!+\!1),\,$ except when $\,n\,$ is odd, when they have gcd $= 2.\,$ The above linear relation between them is simply the Bezout identity for their gcd, considered as a polynomial over $\Bbb Q$ (which can be computed mechanically using the extended Euclidean algorithm). Though this approach is not as efficient as using modular arithmetic, it highlights an interesting viewpoint that often proves useful: often properties of integers (numbers) are special cases of properties of polynomials (functions).

Solution 3:

Hint: What are the only squares modulo $3$? Put another way, look at the expression $n^{2}+1$ modulo $3$. What is true of $n^{2} \pmod 3$ for any $n \in \mathbb{N}$?

Solution 4:

If $3$ divides $n^2+1$ then it must have solution modulo $3$. But clearly

$0^2+1\equiv 1 \pmod 3$

$1^2+1 \equiv 2 \pmod 3$

$2^2+1 \equiv 5 \equiv 2 \pmod 3$

Otherwise put $n=3k,3k+1,3k+2$ and see that $3$ never divides it