1

The following is my approach:

Since we're using contraposition approach, we assume that $n + 2$ is a perfect square.

We assume that $√(n+2)=p$

Then $p$ may be even or odd

Let $p = 2q$ (for even) or $p = 2q+1$ (for odd)

Now $p² = (2q)^2 = 4q^2 = 2k, k = 2q^2 ∈ Z$ OR

$p² = (2q+1)^2 = (4q²+4q+1) = 2(2q^2+2q) + 1 = 2k+1, k ∈ Z$ ;

So, any perfect square will be one of the forms $2k$ or $2k + 1$

If $n + 2 = 2k$, then $n = 2k – 2$ which is not a perfect square

If $n + 2 = 2k + 1$, then $n = 2k -1$ which is not a perfect square either

In whatever form a perfect square $n + 2$ may be, $n$ is not perfect square.

I was told that my approach is wrong and that odd and even does not matter in this context. May I please some feedbacks of the public?

Nigel Ang
  • 11
  • $n =2(3) -2 = 4 = 2^2$ is a perfect square. And $n=2(5)-1 = 9 = 3^2$ is also a perfect square. – Robert Lee Jul 26 '21 at 04:41
  • 1
    The problem is that your forms for $n$ are the same as your forms for $n + 2,$ if you just let $k_2 = k - 1.$

    I think this might be harder to see here because you're using the same variable to represent two different values: one which is arbitrary (the $2k$ and $2k + 1$ forms for your squares) and one which is fixed based on your choice of $n.$ ($n = 2k$ or $2k + 1$)

     – Stephen Donovan Jul 26 '21 at 04:48
  • @RobertLee You're substituting 3 into the arbitrary equation. In that sense, 3 would be $k$ instead of $n$ and $k ≠ n$ – Nigel Ang Jul 26 '21 at 04:52
  • @StephenDonovan Alright thanks for your clarification – Nigel Ang Jul 26 '21 at 04:52
  • 3
    Perhaps you'd be better off starting with $n=a^2$ and $n+2=b^2$ and obtaining an interesting expression for $2$ – Hagen von Eitzen Jul 26 '21 at 05:23
  • I think a better way to see this is like this : If $p$ is even, $n = p^2 = 4k$, with some $k \in \mathbb{Z}$. If $p$ is odd, then $n = p^2 = 4k + 1$ for some $k \in \mathbb{Z}$. From there, you can easily find that if both $n$ and $n+2$ are perfect squares, then there follows a contradiction. – Romain Echaniz Jul 26 '21 at 10:40

0 Answers0