MHB Sufficient Proof: $4 \nmid n^2 - 2$ for $r=0$

[tmt1]

I would like to prove that this is incorrect:

$\exists x \in \Bbb{Z}$ such that $ 4 | n^2 - 2$

I can use the quotient remainder theorem,

$n = dq + r$ where $ 0 <= r < d $ and $ d = 4$

For the case $ r = 0$ is this sufficient proof?

$n = 4q $ and $4 | n^2 - 2$ thus $4 | 16q^2 - 2$

then $4 | 4(4q^2) + 2$

which can't be true, $\therefore $ for the case $ r = 0$, $4 \nmid n^2 - 2$

Is this sufficient for the case $r = 0$? (I can figure out the rest of the cases from here)

 

[Greg]

There's no need to use the "quotient remainder theorem" (unless you've been directed to do so).

It's easily proved that all perfect squares are either 0 or 1 mod 4, thus n² - 2 is 2 or 3 mod 4.