Prove that for all integers $n$, $3$ does not divide $n^2-5$ using modular arithmetic.

Question

I am having trouble proving that for all integers $n,\ 3$ does not divide $n^2-5$ using modular arithmetic. I know that $3\not\mid n^2-5$ means $n^2\not\equiv 5\pmod 3$. But I'm not sure how to start the proof.

Answer 1

The only squares mod $3$ are $0$ and $1$. Try this yourself: take the possible remainders mod $3$, square them, and then mod $3$ and see what you get. Now $-5 \equiv 1$ mod $3$. But then $n^2-5 \equiv n^2 + 1$ mod $3$, which can only take values $1$ or $2$ mod $3$. But because this remainder is never $0$, it cannot be divisible by $3$.

Answer 2

Case 1: $$n\equiv 0\pmod 3\Rightarrow n^2\equiv 0\pmod 3\Rightarrow n^2-5\equiv -5\pmod 3\not\equiv 0\pmod 3$$

Case2: $$n\equiv 1\pmod 3\Rightarrow n^2\equiv 1\pmod 3\Rightarrow n^2-5\equiv -4\pmod 3\not\equiv 0\pmod 3$$

Case 3: $$n\equiv 2\pmod 3\Rightarrow n^2\equiv 1\pmod 3\Rightarrow n^2-5\equiv -4\pmod 3\not\equiv 0\pmod 3$$

Answer 3

Suppose there is an $a$ such that $3|(a^2-5)$.

Now, take $n=a-3$.

And see $(a^2-5)-(n^2-5)=3(2a-3)$. This means if $3|(n^2-5)$ for some $n \in \mathbb Z$ then all $(n-3k)^2-5, k \in \mathbb Z$ is divisible by $3$.

If there is such $n$ anywhere far apart, that would be near us also. As we can't find any such $n$ in between $0$ and $3$, it's nowhere.