Proving there's no $n \in \mathbb{Z}$ such that $n^2 = 2$ I feel that my attempt is not what is expected in a Real Analysis course and that there's something I should incorporate from the Algebraic properties of Real numbers. My hunch is that there's better proof with induction, but was not able to think of any and could be wrong.
My attempt:
$$n^2 = 2 $$
$$n^2 = 1 + 1 $$
$$n^2+(-1)=1+1+(-1) $$
$$n^2-1=1+(1+(-1)) $$
$$n^2-1=1$$
Case $n=0$: then $-1=1$
Case $n \neq 0$: then $n^2 \geq 1$ and $n^2-1 \neq 1$

$(-n)^2=n^2$ so we can restrict ourselves to $n\ge 0$
For $n\le 1$ then multiplying by $n$ we get $n^2\le n\le 1$
For $n\ge 2$ then multiplying by $n$ we get $n^2\ge 2n\ge 4$
Therefore neither $n^2=2$ nor $n^2=3$ are reachable.
