Show that n+2 is not a perfect square

[opt!kal]

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:

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!

 

[Goldenwind]

-3^2 = 9
-2^2 = 4
-1^2 = 1
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
And so on.

Regardless of what integer you're squaring, no two perfect squares are a distance of 2 apart. It is 1, 3, 5, 7, etc.

 

[HallsofIvy]

-3^2 = 9
-2^2 = 4
-1^2 = 1
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
And so on.

Regardless of what integer you're squaring, no two perfect squares are a distance of 2 apart. It is 1, 3, 5, 7, etc.

But how would you prove that statement?

 

[HallsofIvy]

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:

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!

You almost have it. Saying that k²+ 2= p² is the same as saying p²- k²= 2. Since the left side is the difference of two squares, that is the same as (p- k)(p+k)= 2. That is, p-k and p+k are integer factors of 2. But 2 only has 1, -1, 2, and -2 as integer factors. Can you finish it from there?

 

[opt!kal]

Yeah I think I see what you're saying. So basically for (k + p)(k - p) = 2 to hold, k and p can only be -2, 2, -1, 1. However, plugging each one in, you see that they will not equal to 2, so that means that the assumption that n + 2 is a perfect square does not hold? Does that seem right? Thanks in advance!

 

[Tedjn]

No, p and k do not need to be -2, 2, -1, or 1. Only p+k and p-k need to be -2, 2, -1, 1.

 

[HallsofIvy]

Yeah I think I see what you're saying. So basically for (k + p)(k - p) = 2 to hold, k and p can only be -2, 2, -1, 1. However, plugging each one in, you see that they will not equal to 2, so that means that the assumption that n + 2 is a perfect square does not hold? Does that seem right? Thanks in advance!

No, p and k do not need to be -2, 2, -1, or 1. Only p+k and p-k need to be -2, 2, -1, 1.

Exactly. And since k and p are positive integers, what can they be if p+ k and p- k are -2, 2, -1, or 1?

 

[opt!kal]

Hrmmm, I think I see what's you're trying to say... So I set p + k = 2 and p - k = 1 and solve right? then do p + k = 1 and p - k = 2...and do the same thing for the negative cases, and solve for p and k, which I don't think will end up as integers...and since an integer a is a perfect square if there is an integer b such that a = b^2, a contradiction occurs? (sorry for the late reply, just got bogged down by other work) Thanks in advance!