- #1
opt!kal
- 19
- 0
Hi there,
I've been having a lot of trouble with this particular proof lately, and I just do not know how to finish it up:
Show that if n is a perfect square, then n+2 is not a perfect square.(show by contradiction)
none that I know of
Since its a proof by contradiction, I know that I can rewrite it in the form: p ^ not q.
So I have: n is a perfect square, and n+2 is a perfect square.
Since I assume that then I have that n = k^2 and n+2 = p^2
So then I'll have (k^2) + 2 = p^2
At this point I basically get confused. I just have no idea on how to proceed from here, my only guess is that I should do two cases, one where k is even and one where k is odd, and show for each case that p^2 will end up like the case, while p will end up as the opposite (i.e. if I do the even case, then try to show that p will end up odd). But I have absolutely no idea on how to accomplish that. Any help would be greatly appreciated!
I've been having a lot of trouble with this particular proof lately, and I just do not know how to finish it up:
Homework Statement
Show that if n is a perfect square, then n+2 is not a perfect square.(show by contradiction)
Homework Equations
none that I know of
The Attempt at a Solution
Since its a proof by contradiction, I know that I can rewrite it in the form: p ^ not q.
So I have: n is a perfect square, and n+2 is a perfect square.
Since I assume that then I have that n = k^2 and n+2 = p^2
So then I'll have (k^2) + 2 = p^2
At this point I basically get confused. I just have no idea on how to proceed from here, my only guess is that I should do two cases, one where k is even and one where k is odd, and show for each case that p^2 will end up like the case, while p will end up as the opposite (i.e. if I do the even case, then try to show that p will end up odd). But I have absolutely no idea on how to accomplish that. Any help would be greatly appreciated!