If $n$ is a perfect square, then $n+2$ is not. Valid proof by contradiction?

Question

I understand this proposition has been proven many times across this forum, but my first attempt at a proof led me to a strategy that I haven't seen anywhere. I'm curious if it is a valid approach.

Let $n = k^2, k \in \mathbb{Z}$. Assume for contradiction that $n+2 = j^2, j \in \mathbb{Z}$.

$n=k^2 \Rightarrow k^2 +2 = j^2 \Rightarrow \sqrt{k^2 + 2} = j$

$j \notin \mathbb{Z}$, which contradicts assumption. Therefore, $n+2$ is not equal to a perfect square.

Did I go wrong anywhere here?

Answer 1

As others commented, how do you know with $k\in\mathbb Z$ that $j=\sqrt{k^2+2}\not\in\mathbb Z$?

Here's a proof:

We may assume without loss of generality that $k\ge0$.

For $k\ge1$, $(k+1)^2=k^2+2k+1>k^2+2=j^2$, so $k<j<k+1$, so $j\not\in\mathbb Z$.

For $k=0, j=\sqrt{k^2+2}$ is not an integer either.

Answer 2

Well, you said nothing wrong, but your jumping from $\sqrt{k^2 + 2} = j$ to the notion that $j$ is not an integer...that's equivalent to the original problem. How do you know that the square root isn't an integer? Lots of square roots are, like $\sqrt{64}$ and $\sqrt{16}$, etc.