# No natural numbers that satisfy x^2 - y^2 = 2

• lisamane
In summary: I'm sorry, I don't really know how to go about anyone's suggestions. So from the given equation, I am trying to prove that there are no natural numbers x and y that satisfy the equation x^2 - y^2 = 2. In summary, the conversation discusses attempting to prove that there are no natural numbers that satisfy the equation x^2 - y^2 = 2. The method of proof by contradiction is suggested, with the assumption that x and y are rational numbers. However, it is pointed out that the equation is for natural numbers, not rational numbers. The conversation then moves on to discussing possible cases, including using the factored form (x-y)(x+y)=2, but it is
lisamane
I'm trying to prove that there are no natural numbers x and y that satisfy the equation x^2 - y^2 = 2.

I tried to solve it by contradiction and so I assume that x and y are rational numbers and both x and y can be written in the form (a/b) where it's in its simplest form and a and b are both integers and b isn't equal to zero.

Therefore let x = (a/b) and y = (c/d)

Therefore (a^2/b^2) - (c^2/d^2) = 2

Then multiply by (b^2)(d^2)

(a^2)(d^2) - (c^2)(b^2) = 2(b^2)(d^2)

I don't know how to go further from here or if I am even on the right track.

I'd say you are not on the right track because you are trying to do it for rational numbers and the question was about natural numbers. (Which are the ornery numbers 1, 2, 3, and you probably know others. )

I would think about the factored form (x-y)(x+y)=2.

"Ornery" numbers?

Dick said:
I would think about the factored form (x-y)(x+y)=2.

So using (x-y)(x+y) = 2 I thought of:

if x and y are equal then left hand side is equal to zero (contradiction)

if y is greater than x
I thought that the left hand side would always be negative but this won't be the case as say
x = -3 and y = 1 , then left hand side is equal to 8 which is not negative, so I can't really use this case.

Is this more on the correct track?

lisamane said:
So using (x-y)(x+y) = 2 I thought of:

if x and y are equal then left hand side is equal to zero (contradiction)

if y is greater than x
I thought that the left hand side would always be negative but this won't be the case as say
x = -3 and y = 1 , then left hand side is equal to 8 which is not negative, so I can't really use this case.

Is this more on the correct track?

Sure. But if x and y are natural numbers then x>0 and y>0 and they are integers. If (x+y)>2 then why can't that work? So the only possibility would be x=1 and y=1. Does that work?

Last edited:
lisamane said:
x = -3 and y = 1

Let me say again I think you need to look up the definition of natural numbers, -3 isn't a natural number. http://en.wikipedia.org/wiki/Natural_number

I don't think you need to use the factorisation. To get 2 you need x>y and from there it is very simple.

HallsofIvy said:
"Ornery" numbers?

I have used them in many unpublished calculations.

Bah, I forgot that natural numbers aren't negative. Would you include zero as a natural?

So from (x-y)(x+y) = 2,

Case 1: ( x = y )
Left side = (0)(x.x) = 0 ≠ 2

Case 2: ( y > x )
Left side is negative integer multiplied by positive integer which is equal to a negative integer which ≠ 2

Case 3: ( x > y )
This is what I'm stuck on but I tried doing it like this.

When both x and y are less than or equal to 2:
y = 1, x = 2: Left side = (1)(3) = 3 ≠ 2
When both greater than 2:
then let x = p + 2 and y = q + 2
(p - q)(p + q + 4) = p.p - p.q + p.q - q.q + 4p + 4q
= p.p - q.q + 4(p+q)
I was trying to show that when they are both greater than two than the result will be greater than 2 but I don't know how to.
These are the only three possible cases?

I say again forget the factorisation.

Let y = x + a where a>1

Later, once you have got it for natural numbers you may well be able to state and prove something for a wider class of numbers - that is what Polya in 'How to Solve It' would recommend.

lisamane said:
So from (x-y)(x+y) = 2,

Case 1: ( x = y )
Left side = (0)(x.x) = 0 ≠ 2

Case 2: ( y > x )
Left side is negative integer multiplied by positive integer which is equal to a negative integer which ≠ 2

Case 3: ( x > y )
This is what I'm stuck on but I tried doing it like this.

When both x and y are less than or equal to 2:
y = 1, x = 2: Left side = (1)(3) = 3 ≠ 2
When both greater than 2:
then let x = p + 2 and y = q + 2
(p - q)(p + q + 4) = p.p - p.q + p.q - q.q + 4p + 4q
= p.p - q.q + 4(p+q)

I'm sorry, I don't really know how to go about anyone's suggestions so continuing with Case 3 I thought since the last is p^2 - q^2 + 4(p+q) then from the equation p^2 - y^2 = 2 so the equation becomes 2 + 4(p + q) and therefore since all positive, all are greater than two unless p and q. Doesn't that work?

Except that when I use American expressions I have certainly heard they seem to be not understood it was in my mind all along to say 'Be a country girl!' meaning I think you are being unnecessarily complicated.

Taking 1 step of my last suggestion you really can't do anything with

(x + a)2 - x2 = 2 ?

epenguin said:
Except that when I use American expressions I have certainly heard they seem to be not understood it was in my mind all along to say 'Be a country girl!' meaning I think you are being unnecessarily complicated.

Taking 1 step of my last suggestion you really can't do anything with

(x + a)2 - x2 = 2 ?

Earlier you said let y = x + a which if you put in the equation it should be

x2 - (x + a)2 = 2

How did it come to (x + a)2 - x2 = 2 ?

A so I did, well I have done more than hint so you will have to supply what I should have said that will make it come to something like that, get the idea, nothing complicated.

lisamane said:
I'm trying to prove that there are no natural numbers x and y that satisfy the equation x^2 - y^2 = 2.

As an alternative, I would show that it isn't true for any integers, and then

(The following method is not meant to be shorter than others already mentioned.)

You could prove by contradiction each of four cases to complete it:

$$1st \ case: \ \ \ \ x \ \ even, \ \ y \ \ even$$

$$2nd \ case: \ \ \ \ x \ \ even, \ \ y \ \ odd$$

$$3rd \ case: \ \ \ \ x \ \ odd, \ \ y \ \ even$$

$$4th \ case: \ \ \ \ x \ \ odd, \ \ y \ \ odd$$

-----------------------------------------------------

For example, let m, n belong to the set of integers

For instance, the second case could be:

Then x = 2m and y = 2n + 1.

Substitute these into the original equation and see if
you can arrive at a contradiction.

Look at it this way: $x^2 - y^2 = (x-y)(x+y)$, right? So suppose that x and y ARE two natural numbers whose product is 2, then we have $(x-y)(x+y) = 2$. There are only two natural numbers whose product is 2, namely the numbers 1 and 2.

So either $(x-y) = 1$ and $(x+y) = 2$ OR $(x-y) = 2$ and $(x+y) = 1$.

Solve both those systems for x and y to get a contradiction.

## What is the definition of a natural number?

A natural number is a positive integer (1,2,3,4...) that is used for counting and ordering objects.

## What does the equation x^2 - y^2 = 2 represent?

This equation represents a difference of squares, where the difference between two perfect squares is equal to 2.

## Why can't there be any natural numbers that satisfy this equation?

This is because the difference of two perfect squares can never equal to 2, as the squares of any two natural numbers will always be at least one apart.

## What is the mathematical proof for this statement?

The proof for this statement is based on the concept of parity, where both x and y must be either both even or both odd for the equation to be satisfied. However, since the difference of two perfect squares can never equal 2, there are no possible combinations of even or odd numbers that can satisfy this equation.

## Are there any other types of numbers that can satisfy this equation?

No, this equation can only be satisfied by irrational numbers such as √2. Rational numbers and other types of numbers cannot satisfy this equation.

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