# Show that n+2 is not a perfect square

• opt!kal
In summary, the conversation discusses a proof by contradiction that shows if n is a perfect square, then n+2 is not a perfect square. The conversation goes through the steps of the proof and concludes by finding a contradiction when trying to solve for p and k. This proves the initial statement to be true.
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!

-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.

Goldenwind said:
-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?

opt!kal said:
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?

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.

opt!kal said:
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 said:
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?

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!

## 1. What does it mean for a number to be a perfect square?

A perfect square is a number that is the result of multiplying a number by itself. For example, 4 is a perfect square because it is the result of 2 multiplied by 2.

## 2. How do you show that n+2 is not a perfect square?

To show that n+2 is not a perfect square, we need to prove that it cannot be written as the product of two identical numbers. We can do this by showing that the number has a remainder when divided by any number less than it.

## 3. Can you provide an example of a number that is not a perfect square?

Yes, any odd number is not a perfect square. For example, 7 is not a perfect square because it cannot be written as the product of two identical numbers.

## 4. Is there a specific formula for determining if a number is a perfect square?

Yes, there is a formula called the square root algorithm that can be used to determine if a number is a perfect square. It involves finding the square root of the number and checking if it is a whole number.

## 5. Why is it important to show that n+2 is not a perfect square?

Showing that n+2 is not a perfect square is important because it helps us understand the properties of numbers and their relationships. It also allows us to make more accurate calculations and predictions in various fields, such as mathematics, science, and engineering.

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