<s>[INST] Proof: $n^2 - 2$ is not divisible by 4 I tried to prove that $n^2 - 2$ is not divisible by 4 via proof by contradiction.  Does this look right?
Suppose $n^2 - 2$ is divisible by $4$.  Then:
$n^2 - 2 = 4g$, $g \in \mathbb{Z}$.
$n^2 = 4g + 2$.
Consider the case where $n$ is even.
$(2x)^2 = 4g + 2$, $x \in \mathbb{Z}$.  
$4x^2 = 4g + 2$.  
$4s = 4g + 2$, $s = x^2, s \in \mathbb{Z}$ as integers are closed under multiplication.  
$2s = 2g + 1$
$2s$ is even, and $2g + 1$ is odd (by definition of even/odd numbers).  An even number cannot equal an odd number, so we have a contradiction.
Consider the case where $n$ is odd.
$(2x + 1)^2 = 4g + 2$, $x \in \mathbb{Z}$
$4x^2 + 4x + 1 = 4g + 2$
$4x^2 + 4x = 4g + 1$
$4(x^2 + x) = 4g + 1$
$4j = 4g + 1$, $j = x^2 + x, j \in \mathbb{Z}$ as integers are closed under addition
$2d = 2e + 1$, $d = 2j, e = 2g; d, e \in \mathbb{Z}$ as integers are closed under multiplication
$2d$ is even, and $2e + 1$ is odd (by definition of even/odd numbers).  An even number cannot equal an odd number, so we have a contradiction.
As both cases have a contradiction, the original supposition is false, and $n^2 - 2$ is not divisible by $4$.
 [/INST] For odd $n$, $n^2-2$ is odd.
For even $n$, $n^2$ is divisible by $4$, so that $n^2-2$ is not.

By contradiction:
Let $n$ be odd. Then $n^2-2$ is both odd and a multiple of four.
Let $n$ be even. Then $n^2-2$ and $n^2$ are both multiples of $4$, so that $2$ is a multiple of $4$.
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