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?

  • 3
    $\begingroup$ why can you conclude that $j\not \in \mathbb{Z}$? $\endgroup$
    – Phicar
    Dec 18 '19 at 2:08
  • $\begingroup$ how do we know it's not integer. $\endgroup$
    – user645636
    Dec 18 '19 at 2:09

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.


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.


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